Algebra Examples

$(- 7, 134)$ , $(- 3, 10)$ , $(5, 50)$

Step 1

Use the standard form of a quadratic equation $y = a x^{2} + b x + c$ as the starting point for finding the equation through the three points.

$y = a x^{2} + b x + c$

Step 2

Create a system of equations by substituting the $x$ and $y$ values of each point into the standard formula of a quadratic equation to create the three equation system.

$134 = a {(- 7)}^{2} + b (- 7) + c, 10 = a {(- 3)}^{2} + b (- 3) + c, 50 = a {(5)}^{2} + b (5) + c$

Step 3

Solve the system of equations to find the values of $a$ , $b$ , and $c$ .

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Step 3.1

Solve for $c$ in $134 = a {(- 7)}^{2} + b (- 7) + c$ .

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Step 3.1.1

Rewrite the equation as $a {(- 7)}^{2} + b (- 7) + c = 134$ .

$a {(- 7)}^{2} + b (- 7) + c = 134$

$10 = a {(- 3)}^{2} + b (- 3) + c$

$50 = a \cdot 5^{2} + b (5) + c$

Step 3.1.2

Simplify each term.

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Step 3.1.2.1

Raise $- 7$ to the power of $2$ .

$a \cdot 49 + b (- 7) + c = 134$

$10 = a {(- 3)}^{2} + b (- 3) + c$

$50 = a \cdot 5^{2} + b (5) + c$

Step 3.1.2.2

Move $49$ to the left of $a$ .

$49 \cdot a + b (- 7) + c = 134$

$10 = a {(- 3)}^{2} + b (- 3) + c$

$50 = a \cdot 5^{2} + b (5) + c$

Step 3.1.2.3

Move $- 7$ to the left of $b$ .

$49 a - 7 b + c = 134$

$10 = a {(- 3)}^{2} + b (- 3) + c$

$50 = a \cdot 5^{2} + b (5) + c$

$49 a - 7 b + c = 134$

$10 = a {(- 3)}^{2} + b (- 3) + c$

$50 = a \cdot 5^{2} + b (5) + c$

Step 3.1.3

Move all terms not containing $c$ to the right side of the equation.

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Step 3.1.3.1

Subtract $49 a$ from both sides of the equation.

$- 7 b + c = 134 - 49 a$

$10 = a {(- 3)}^{2} + b (- 3) + c$

$50 = a \cdot 5^{2} + b (5) + c$

Step 3.1.3.2

Add $7 b$ to both sides of the equation.

$c = 134 - 49 a + 7 b$

$10 = a {(- 3)}^{2} + b (- 3) + c$

$50 = a \cdot 5^{2} + b (5) + c$

$c = 134 - 49 a + 7 b$

$10 = a {(- 3)}^{2} + b (- 3) + c$

$50 = a \cdot 5^{2} + b (5) + c$

$c = 134 - 49 a + 7 b$

$10 = a {(- 3)}^{2} + b (- 3) + c$

$50 = a \cdot 5^{2} + b (5) + c$

Step 3.2

Replace all occurrences of $c$ with $134 - 49 a + 7 b$ in each equation.

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Step 3.2.1

Replace all occurrences of $c$ in $10 = a {(- 3)}^{2} + b (- 3) + c$ with $134 - 49 a + 7 b$ .

$10 = a {(- 3)}^{2} + b (- 3) + 134 - 49 a + 7 b$

$c = 134 - 49 a + 7 b$

$50 = a \cdot 5^{2} + b (5) + c$

Step 3.2.2

Simplify $10 = a {(- 3)}^{2} + b (- 3) + 134 - 49 a + 7 b$ .

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Step 3.2.2.1

Simplify the left side.

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Step 3.2.2.1.1

Remove parentheses.

$10 = a {(- 3)}^{2} + b (- 3) + 134 - 49 a + 7 b$

$c = 134 - 49 a + 7 b$

$50 = a \cdot 5^{2} + b (5) + c$

$10 = a {(- 3)}^{2} + b (- 3) + 134 - 49 a + 7 b$

$c = 134 - 49 a + 7 b$

$50 = a \cdot 5^{2} + b (5) + c$

Step 3.2.2.2

Simplify the right side.

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Step 3.2.2.2.1

Simplify $a {(- 3)}^{2} + b (- 3) + 134 - 49 a + 7 b$ .

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Step 3.2.2.2.1.1

Simplify each term.

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Step 3.2.2.2.1.1.1

Raise $- 3$ to the power of $2$ .

$10 = a \cdot 9 + b (- 3) + 134 - 49 a + 7 b$

$c = 134 - 49 a + 7 b$

$50 = a \cdot 5^{2} + b (5) + c$

Step 3.2.2.2.1.1.2

Move $9$ to the left of $a$ .

$10 = 9 \cdot a + b (- 3) + 134 - 49 a + 7 b$

$c = 134 - 49 a + 7 b$

$50 = a \cdot 5^{2} + b (5) + c$

Step 3.2.2.2.1.1.3

Move $- 3$ to the left of $b$ .

$10 = 9 a - 3 b + 134 - 49 a + 7 b$

$c = 134 - 49 a + 7 b$

$50 = a \cdot 5^{2} + b (5) + c$

$10 = 9 a - 3 b + 134 - 49 a + 7 b$

$c = 134 - 49 a + 7 b$

$50 = a \cdot 5^{2} + b (5) + c$

Step 3.2.2.2.1.2

Simplify by adding terms.

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Step 3.2.2.2.1.2.1

Subtract $49 a$ from $9 a$ .

$10 = - 40 a - 3 b + 134 + 7 b$

$c = 134 - 49 a + 7 b$

$50 = a \cdot 5^{2} + b (5) + c$

Step 3.2.2.2.1.2.2

Add $- 3 b$ and $7 b$ .

$10 = - 40 a + 4 b + 134$

$c = 134 - 49 a + 7 b$

$50 = a \cdot 5^{2} + b (5) + c$

$10 = - 40 a + 4 b + 134$

$c = 134 - 49 a + 7 b$

$50 = a \cdot 5^{2} + b (5) + c$

$10 = - 40 a + 4 b + 134$

$c = 134 - 49 a + 7 b$

$50 = a \cdot 5^{2} + b (5) + c$

$10 = - 40 a + 4 b + 134$

$c = 134 - 49 a + 7 b$

$50 = a \cdot 5^{2} + b (5) + c$

$10 = - 40 a + 4 b + 134$

$c = 134 - 49 a + 7 b$

$50 = a \cdot 5^{2} + b (5) + c$

Step 3.2.3

Replace all occurrences of $c$ in $50 = a \cdot 5^{2} + b (5) + c$ with $134 - 49 a + 7 b$ .

$50 = a \cdot 5^{2} + b (5) + 134 - 49 a + 7 b$

$10 = - 40 a + 4 b + 134$

$c = 134 - 49 a + 7 b$

Step 3.2.4

Simplify $50 = a \cdot 5^{2} + b (5) + 134 - 49 a + 7 b$ .

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Step 3.2.4.1

Simplify the left side.

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Step 3.2.4.1.1

Remove parentheses.

$50 = a \cdot 5^{2} + b (5) + 134 - 49 a + 7 b$

$10 = - 40 a + 4 b + 134$

$c = 134 - 49 a + 7 b$

$50 = a \cdot 5^{2} + b (5) + 134 - 49 a + 7 b$

$10 = - 40 a + 4 b + 134$

$c = 134 - 49 a + 7 b$

Step 3.2.4.2

Simplify the right side.

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Step 3.2.4.2.1

Simplify $a \cdot 5^{2} + b (5) + 134 - 49 a + 7 b$ .

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Step 3.2.4.2.1.1

Simplify each term.

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Step 3.2.4.2.1.1.1

Raise $5$ to the power of $2$ .

$50 = a \cdot 25 + b (5) + 134 - 49 a + 7 b$

$10 = - 40 a + 4 b + 134$

$c = 134 - 49 a + 7 b$

Step 3.2.4.2.1.1.2

Move $25$ to the left of $a$ .

$50 = 25 \cdot a + b (5) + 134 - 49 a + 7 b$

$10 = - 40 a + 4 b + 134$

$c = 134 - 49 a + 7 b$

Step 3.2.4.2.1.1.3

Move $5$ to the left of $b$ .

$50 = 25 a + 5 b + 134 - 49 a + 7 b$

$10 = - 40 a + 4 b + 134$

$c = 134 - 49 a + 7 b$

$50 = 25 a + 5 b + 134 - 49 a + 7 b$

$10 = - 40 a + 4 b + 134$

$c = 134 - 49 a + 7 b$

Step 3.2.4.2.1.2

Simplify by adding terms.

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Step 3.2.4.2.1.2.1

Subtract $49 a$ from $25 a$ .

$50 = - 24 a + 5 b + 134 + 7 b$

$10 = - 40 a + 4 b + 134$

$c = 134 - 49 a + 7 b$

Step 3.2.4.2.1.2.2

Add $5 b$ and $7 b$ .

$50 = - 24 a + 12 b + 134$

$10 = - 40 a + 4 b + 134$

$c = 134 - 49 a + 7 b$

$50 = - 24 a + 12 b + 134$

$10 = - 40 a + 4 b + 134$

$c = 134 - 49 a + 7 b$

$50 = - 24 a + 12 b + 134$

$10 = - 40 a + 4 b + 134$

$c = 134 - 49 a + 7 b$

$50 = - 24 a + 12 b + 134$

$10 = - 40 a + 4 b + 134$

$c = 134 - 49 a + 7 b$

$50 = - 24 a + 12 b + 134$

$10 = - 40 a + 4 b + 134$

$c = 134 - 49 a + 7 b$

$50 = - 24 a + 12 b + 134$

$10 = - 40 a + 4 b + 134$

$c = 134 - 49 a + 7 b$

Step 3.3

Solve for $a$ in $50 = - 24 a + 12 b + 134$ .

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Step 3.3.1

Rewrite the equation as $- 24 a + 12 b + 134 = 50$ .

$- 24 a + 12 b + 134 = 50$

$10 = - 40 a + 4 b + 134$

$c = 134 - 49 a + 7 b$

Step 3.3.2

Move all terms not containing $a$ to the right side of the equation.

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Step 3.3.2.1

Subtract $12 b$ from both sides of the equation.

$- 24 a + 134 = 50 - 12 b$

$10 = - 40 a + 4 b + 134$

$c = 134 - 49 a + 7 b$

Step 3.3.2.2

Subtract $134$ from both sides of the equation.

$- 24 a = 50 - 12 b - 134$

$10 = - 40 a + 4 b + 134$

$c = 134 - 49 a + 7 b$

Step 3.3.2.3

Subtract $134$ from $50$ .

$- 24 a = - 12 b - 84$

$10 = - 40 a + 4 b + 134$

$c = 134 - 49 a + 7 b$

$- 24 a = - 12 b - 84$

$10 = - 40 a + 4 b + 134$

$c = 134 - 49 a + 7 b$

Step 3.3.3

Divide each term in $- 24 a = - 12 b - 84$ by $- 24$ and simplify.

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Step 3.3.3.1

Divide each term in $- 24 a = - 12 b - 84$ by $- 24$ .

$\frac{- 24 a}{- 24} = \frac{- 12 b}{- 24} + \frac{- 84}{- 24}$

$10 = - 40 a + 4 b + 134$

$c = 134 - 49 a + 7 b$

Step 3.3.3.2

Simplify the left side.

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Step 3.3.3.2.1

Cancel the common factor of $- 24$ .

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Step 3.3.3.2.1.1

Cancel the common factor.

$\frac{- 24 a}{- 24} = \frac{- 12 b}{- 24} + \frac{- 84}{- 24}$

$10 = - 40 a + 4 b + 134$

$c = 134 - 49 a + 7 b$

Step 3.3.3.2.1.2

Divide $a$ by $1$ .

$a = \frac{- 12 b}{- 24} + \frac{- 84}{- 24}$

$10 = - 40 a + 4 b + 134$

$c = 134 - 49 a + 7 b$

$a = \frac{- 12 b}{- 24} + \frac{- 84}{- 24}$

$10 = - 40 a + 4 b + 134$

$c = 134 - 49 a + 7 b$

$a = \frac{- 12 b}{- 24} + \frac{- 84}{- 24}$

$10 = - 40 a + 4 b + 134$

$c = 134 - 49 a + 7 b$

Step 3.3.3.3

Simplify the right side.

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Step 3.3.3.3.1

Simplify each term.

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Step 3.3.3.3.1.1

Cancel the common factor of $- 12$ and $- 24$ .

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Step 3.3.3.3.1.1.1

Factor $- 12$ out of $- 12 b$ .

$a = \frac{- 12 b}{- 24} + \frac{- 84}{- 24}$

$10 = - 40 a + 4 b + 134$

$c = 134 - 49 a + 7 b$

Step 3.3.3.3.1.1.2

Cancel the common factors.

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Step 3.3.3.3.1.1.2.1

Factor $- 12$ out of $- 24$ .

$a = \frac{- 12 b}{- 12 \cdot 2} + \frac{- 84}{- 24}$

$10 = - 40 a + 4 b + 134$

$c = 134 - 49 a + 7 b$

Step 3.3.3.3.1.1.2.2

Cancel the common factor.

$a = \frac{- 12 b}{- 12 \cdot 2} + \frac{- 84}{- 24}$

$10 = - 40 a + 4 b + 134$

$c = 134 - 49 a + 7 b$

Step 3.3.3.3.1.1.2.3

Rewrite the expression.

$a = \frac{b}{2} + \frac{- 84}{- 24}$

$10 = - 40 a + 4 b + 134$

$c = 134 - 49 a + 7 b$

$a = \frac{b}{2} + \frac{- 84}{- 24}$

$10 = - 40 a + 4 b + 134$

$c = 134 - 49 a + 7 b$

$a = \frac{b}{2} + \frac{- 84}{- 24}$

$10 = - 40 a + 4 b + 134$

$c = 134 - 49 a + 7 b$

Step 3.3.3.3.1.2

Cancel the common factor of $- 84$ and $- 24$ .

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Step 3.3.3.3.1.2.1

Factor $- 12$ out of $- 84$ .

$a = \frac{b}{2} + \frac{- 12 \cdot 7}{- 24}$

$10 = - 40 a + 4 b + 134$

$c = 134 - 49 a + 7 b$

Step 3.3.3.3.1.2.2

Cancel the common factors.

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Step 3.3.3.3.1.2.2.1

Factor $- 12$ out of $- 24$ .

$a = \frac{b}{2} + \frac{- 12 \cdot 7}{- 12 \cdot 2}$

$10 = - 40 a + 4 b + 134$

$c = 134 - 49 a + 7 b$

Step 3.3.3.3.1.2.2.2

Cancel the common factor.

$a = \frac{b}{2} + \frac{- 12 \cdot 7}{- 12 \cdot 2}$

$10 = - 40 a + 4 b + 134$

$c = 134 - 49 a + 7 b$

Step 3.3.3.3.1.2.2.3

Rewrite the expression.

$a = \frac{b}{2} + \frac{7}{2}$

$10 = - 40 a + 4 b + 134$

$c = 134 - 49 a + 7 b$

$a = \frac{b}{2} + \frac{7}{2}$

$10 = - 40 a + 4 b + 134$

$c = 134 - 49 a + 7 b$

$a = \frac{b}{2} + \frac{7}{2}$

$10 = - 40 a + 4 b + 134$

$c = 134 - 49 a + 7 b$

$a = \frac{b}{2} + \frac{7}{2}$

$10 = - 40 a + 4 b + 134$

$c = 134 - 49 a + 7 b$

$a = \frac{b}{2} + \frac{7}{2}$

$10 = - 40 a + 4 b + 134$

$c = 134 - 49 a + 7 b$

$a = \frac{b}{2} + \frac{7}{2}$

$10 = - 40 a + 4 b + 134$

$c = 134 - 49 a + 7 b$

$a = \frac{b}{2} + \frac{7}{2}$

$10 = - 40 a + 4 b + 134$

$c = 134 - 49 a + 7 b$

Step 3.4

Replace all occurrences of $a$ with $\frac{b}{2} + \frac{7}{2}$ in each equation.

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Step 3.4.1

Replace all occurrences of $a$ in $10 = - 40 a + 4 b + 134$ with $\frac{b}{2} + \frac{7}{2}$ .

$10 = - 40 (\frac{b}{2} + \frac{7}{2}) + 4 b + 134$

$a = \frac{b}{2} + \frac{7}{2}$

$c = 134 - 49 a + 7 b$

Step 3.4.2

Simplify the right side.

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Step 3.4.2.1

Simplify $- 40 (\frac{b}{2} + \frac{7}{2}) + 4 b + 134$ .

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Step 3.4.2.1.1

Simplify each term.

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Step 3.4.2.1.1.1

Apply the distributive property.

$10 = - 40 \frac{b}{2} - 40 (\frac{7}{2}) + 4 b + 134$

$a = \frac{b}{2} + \frac{7}{2}$

$c = 134 - 49 a + 7 b$

Step 3.4.2.1.1.2

Cancel the common factor of $2$ .

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Step 3.4.2.1.1.2.1

Factor $2$ out of $- 40$ .

$10 = 2 (- 20) (\frac{b}{2}) - 40 (\frac{7}{2}) + 4 b + 134$

$a = \frac{b}{2} + \frac{7}{2}$

$c = 134 - 49 a + 7 b$

Step 3.4.2.1.1.2.2

Cancel the common factor.

$10 = 2 \cdot (- 20 \frac{b}{2}) - 40 (\frac{7}{2}) + 4 b + 134$

$a = \frac{b}{2} + \frac{7}{2}$

$c = 134 - 49 a + 7 b$

Step 3.4.2.1.1.2.3

Rewrite the expression.

$10 = - 20 b - 40 (\frac{7}{2}) + 4 b + 134$

$a = \frac{b}{2} + \frac{7}{2}$

$c = 134 - 49 a + 7 b$

$10 = - 20 b - 40 (\frac{7}{2}) + 4 b + 134$

$a = \frac{b}{2} + \frac{7}{2}$

$c = 134 - 49 a + 7 b$

Step 3.4.2.1.1.3

Cancel the common factor of $2$ .

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Step 3.4.2.1.1.3.1

Factor $2$ out of $- 40$ .

$10 = - 20 b + 2 (- 20) (\frac{7}{2}) + 4 b + 134$

$a = \frac{b}{2} + \frac{7}{2}$

$c = 134 - 49 a + 7 b$

Step 3.4.2.1.1.3.2

Cancel the common factor.

$10 = - 20 b + 2 \cdot (- 20 (\frac{7}{2})) + 4 b + 134$

$a = \frac{b}{2} + \frac{7}{2}$

$c = 134 - 49 a + 7 b$

Step 3.4.2.1.1.3.3

Rewrite the expression.

$10 = - 20 b - 20 \cdot 7 + 4 b + 134$

$a = \frac{b}{2} + \frac{7}{2}$

$c = 134 - 49 a + 7 b$

$10 = - 20 b - 20 \cdot 7 + 4 b + 134$

$a = \frac{b}{2} + \frac{7}{2}$

$c = 134 - 49 a + 7 b$

Step 3.4.2.1.1.4

Multiply $- 20$ by $7$ .

$10 = - 20 b - 140 + 4 b + 134$

$a = \frac{b}{2} + \frac{7}{2}$

$c = 134 - 49 a + 7 b$

$10 = - 20 b - 140 + 4 b + 134$

$a = \frac{b}{2} + \frac{7}{2}$

$c = 134 - 49 a + 7 b$

Step 3.4.2.1.2

Simplify by adding terms.

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Step 3.4.2.1.2.1

Add $- 20 b$ and $4 b$ .

$10 = - 16 b - 140 + 134$

$a = \frac{b}{2} + \frac{7}{2}$

$c = 134 - 49 a + 7 b$

Step 3.4.2.1.2.2

Add $- 140$ and $134$ .

$10 = - 16 b - 6$

$a = \frac{b}{2} + \frac{7}{2}$

$c = 134 - 49 a + 7 b$

$10 = - 16 b - 6$

$a = \frac{b}{2} + \frac{7}{2}$

$c = 134 - 49 a + 7 b$

$10 = - 16 b - 6$

$a = \frac{b}{2} + \frac{7}{2}$

$c = 134 - 49 a + 7 b$

$10 = - 16 b - 6$

$a = \frac{b}{2} + \frac{7}{2}$

$c = 134 - 49 a + 7 b$

Step 3.4.3

Replace all occurrences of $a$ in $c = 134 - 49 a + 7 b$ with $\frac{b}{2} + \frac{7}{2}$ .

$c = 134 - 49 (\frac{b}{2} + \frac{7}{2}) + 7 b$

$10 = - 16 b - 6$

$a = \frac{b}{2} + \frac{7}{2}$

Step 3.4.4

Simplify the right side.

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Step 3.4.4.1

Simplify $134 - 49 (\frac{b}{2} + \frac{7}{2}) + 7 b$ .

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Step 3.4.4.1.1

Simplify each term.

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Step 3.4.4.1.1.1

Apply the distributive property.

$c = 134 - 49 \frac{b}{2} - 49 (\frac{7}{2}) + 7 b$

$10 = - 16 b - 6$

$a = \frac{b}{2} + \frac{7}{2}$

Step 3.4.4.1.1.2

Combine $- 49$ and $\frac{b}{2}$ .

$c = 134 + \frac{- 49 b}{2} - 49 (\frac{7}{2}) + 7 b$

$10 = - 16 b - 6$

$a = \frac{b}{2} + \frac{7}{2}$

Step 3.4.4.1.1.3

Multiply $- 49 (\frac{7}{2})$ .

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Step 3.4.4.1.1.3.1

Combine $- 49$ and $\frac{7}{2}$ .

$c = 134 + \frac{- 49 b}{2} + \frac{- 49 \cdot 7}{2} + 7 b$

$10 = - 16 b - 6$

$a = \frac{b}{2} + \frac{7}{2}$

Step 3.4.4.1.1.3.2

Multiply $- 49$ by $7$ .

$c = 134 + \frac{- 49 b}{2} + \frac{- 343}{2} + 7 b$

$10 = - 16 b - 6$

$a = \frac{b}{2} + \frac{7}{2}$

$c = 134 + \frac{- 49 b}{2} + \frac{- 343}{2} + 7 b$

$10 = - 16 b - 6$

$a = \frac{b}{2} + \frac{7}{2}$

Step 3.4.4.1.1.4

Simplify each term.

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Step 3.4.4.1.1.4.1

Move the negative in front of the fraction.

$c = 134 - \frac{(49) b}{2} + \frac{- 343}{2} + 7 b$

$10 = - 16 b - 6$

$a = \frac{b}{2} + \frac{7}{2}$

Step 3.4.4.1.1.4.2

Move the negative in front of the fraction.

$c = 134 - \frac{49 b}{2} - \frac{343}{2} + 7 b$

$10 = - 16 b - 6$

$a = \frac{b}{2} + \frac{7}{2}$

$c = 134 - \frac{49 b}{2} - \frac{343}{2} + 7 b$

$10 = - 16 b - 6$

$a = \frac{b}{2} + \frac{7}{2}$

$c = 134 - \frac{49 b}{2} - \frac{343}{2} + 7 b$

$10 = - 16 b - 6$

$a = \frac{b}{2} + \frac{7}{2}$

Step 3.4.4.1.2

To write $134$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$c = - \frac{49 b}{2} + 134 \cdot \frac{2}{2} - \frac{343}{2} + 7 b$

$10 = - 16 b - 6$

$a = \frac{b}{2} + \frac{7}{2}$

Step 3.4.4.1.3

Combine $134$ and $\frac{2}{2}$ .

$c = - \frac{49 b}{2} + \frac{134 \cdot 2}{2} - \frac{343}{2} + 7 b$

$10 = - 16 b - 6$

$a = \frac{b}{2} + \frac{7}{2}$

Step 3.4.4.1.4

Combine the numerators over the common denominator.

$c = - \frac{49 b}{2} + \frac{134 \cdot 2 - 343}{2} + 7 b$

$10 = - 16 b - 6$

$a = \frac{b}{2} + \frac{7}{2}$

Step 3.4.4.1.5

Simplify the numerator.

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Step 3.4.4.1.5.1

Multiply $134$ by $2$ .

$c = - \frac{49 b}{2} + \frac{268 - 343}{2} + 7 b$

$10 = - 16 b - 6$

$a = \frac{b}{2} + \frac{7}{2}$

Step 3.4.4.1.5.2

Subtract $343$ from $268$ .

$c = - \frac{49 b}{2} + \frac{- 75}{2} + 7 b$

$10 = - 16 b - 6$

$a = \frac{b}{2} + \frac{7}{2}$

$c = - \frac{49 b}{2} + \frac{- 75}{2} + 7 b$

$10 = - 16 b - 6$

$a = \frac{b}{2} + \frac{7}{2}$

Step 3.4.4.1.6

Move the negative in front of the fraction.

$c = - \frac{49 b}{2} - \frac{75}{2} + 7 b$

$10 = - 16 b - 6$

$a = \frac{b}{2} + \frac{7}{2}$

Step 3.4.4.1.7

To write $7 b$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$c = - \frac{49 b}{2} + 7 b \cdot \frac{2}{2} - \frac{75}{2}$

$10 = - 16 b - 6$

$a = \frac{b}{2} + \frac{7}{2}$

Step 3.4.4.1.8

Combine $7 b$ and $\frac{2}{2}$ .

$c = - \frac{49 b}{2} + \frac{7 b \cdot 2}{2} - \frac{75}{2}$

$10 = - 16 b - 6$

$a = \frac{b}{2} + \frac{7}{2}$

Step 3.4.4.1.9

Combine the numerators over the common denominator.

$c = \frac{- 49 b + 7 b \cdot 2}{2} - \frac{75}{2}$

$10 = - 16 b - 6$

$a = \frac{b}{2} + \frac{7}{2}$

Step 3.4.4.1.10

Combine the numerators over the common denominator.

$c = \frac{- 49 b + 7 b \cdot 2 - 75}{2}$

$10 = - 16 b - 6$

$a = \frac{b}{2} + \frac{7}{2}$

Step 3.4.4.1.11

Multiply $2$ by $7$ .

$c = \frac{- 49 b + 14 b - 75}{2}$

$10 = - 16 b - 6$

$a = \frac{b}{2} + \frac{7}{2}$

Step 3.4.4.1.12

Add $- 49 b$ and $14 b$ .

$c = \frac{- 35 b - 75}{2}$

$10 = - 16 b - 6$

$a = \frac{b}{2} + \frac{7}{2}$

Step 3.4.4.1.13

Factor $5$ out of $- 35 b - 75$ .

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Step 3.4.4.1.13.1

Factor $5$ out of $- 35 b$ .

$c = \frac{5 (- 7 b) - 75}{2}$

$10 = - 16 b - 6$

$a = \frac{b}{2} + \frac{7}{2}$

Step 3.4.4.1.13.2

Factor $5$ out of $- 75$ .

$c = \frac{5 (- 7 b) + 5 (- 15)}{2}$

$10 = - 16 b - 6$

$a = \frac{b}{2} + \frac{7}{2}$

Step 3.4.4.1.13.3

Factor $5$ out of $5 (- 7 b) + 5 (- 15)$ .

$c = \frac{5 (- 7 b - 15)}{2}$

$10 = - 16 b - 6$

$a = \frac{b}{2} + \frac{7}{2}$

$c = \frac{5 (- 7 b - 15)}{2}$

$10 = - 16 b - 6$

$a = \frac{b}{2} + \frac{7}{2}$

Step 3.4.4.1.14

Factor $- 1$ out of $- 7 b$ .

$c = \frac{5 (- (7 b) - 15)}{2}$

$10 = - 16 b - 6$

$a = \frac{b}{2} + \frac{7}{2}$

Step 3.4.4.1.15

Rewrite $- 15$ as $- 1 (15)$ .

$c = \frac{5 (- (7 b) - 1 \cdot 15)}{2}$

$10 = - 16 b - 6$

$a = \frac{b}{2} + \frac{7}{2}$

Step 3.4.4.1.16

Factor $- 1$ out of $- (7 b) - 1 (15)$ .

$c = \frac{5 (- (7 b + 15))}{2}$

$10 = - 16 b - 6$

$a = \frac{b}{2} + \frac{7}{2}$

Step 3.4.4.1.17

Simplify the expression.

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Step 3.4.4.1.17.1

Rewrite $- (7 b + 15)$ as $- 1 (7 b + 15)$ .

$c = \frac{5 (- 1 (7 b + 15))}{2}$

$10 = - 16 b - 6$

$a = \frac{b}{2} + \frac{7}{2}$

Step 3.4.4.1.17.2

Move the negative in front of the fraction.

$c = - \frac{5 (7 b + 15)}{2}$

$10 = - 16 b - 6$

$a = \frac{b}{2} + \frac{7}{2}$

$c = - \frac{5 (7 b + 15)}{2}$

$10 = - 16 b - 6$

$a = \frac{b}{2} + \frac{7}{2}$

$c = - \frac{5 (7 b + 15)}{2}$

$10 = - 16 b - 6$

$a = \frac{b}{2} + \frac{7}{2}$

$c = - \frac{5 (7 b + 15)}{2}$

$10 = - 16 b - 6$

$a = \frac{b}{2} + \frac{7}{2}$

$c = - \frac{5 (7 b + 15)}{2}$

$10 = - 16 b - 6$

$a = \frac{b}{2} + \frac{7}{2}$

Step 3.5

Solve for $b$ in $10 = - 16 b - 6$ .

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Step 3.5.1

Rewrite the equation as $- 16 b - 6 = 10$ .

$- 16 b - 6 = 10$

$c = - \frac{5 (7 b + 15)}{2}$

$a = \frac{b}{2} + \frac{7}{2}$

Step 3.5.2

Move all terms not containing $b$ to the right side of the equation.

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Step 3.5.2.1

Add $6$ to both sides of the equation.

$- 16 b = 10 + 6$

$c = - \frac{5 (7 b + 15)}{2}$

$a = \frac{b}{2} + \frac{7}{2}$

Step 3.5.2.2

Add $10$ and $6$ .

$- 16 b = 16$

$c = - \frac{5 (7 b + 15)}{2}$

$a = \frac{b}{2} + \frac{7}{2}$

$- 16 b = 16$

$c = - \frac{5 (7 b + 15)}{2}$

$a = \frac{b}{2} + \frac{7}{2}$

Step 3.5.3

Divide each term in $- 16 b = 16$ by $- 16$ and simplify.

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Step 3.5.3.1

Divide each term in $- 16 b = 16$ by $- 16$ .

$\frac{- 16 b}{- 16} = \frac{16}{- 16}$

$c = - \frac{5 (7 b + 15)}{2}$

$a = \frac{b}{2} + \frac{7}{2}$

Step 3.5.3.2

Simplify the left side.

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Step 3.5.3.2.1

Cancel the common factor of $- 16$ .

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Step 3.5.3.2.1.1

Cancel the common factor.

$\frac{- 16 b}{- 16} = \frac{16}{- 16}$

$c = - \frac{5 (7 b + 15)}{2}$

$a = \frac{b}{2} + \frac{7}{2}$

Step 3.5.3.2.1.2

Divide $b$ by $1$ .

$b = \frac{16}{- 16}$

$c = - \frac{5 (7 b + 15)}{2}$

$a = \frac{b}{2} + \frac{7}{2}$

$b = \frac{16}{- 16}$

$c = - \frac{5 (7 b + 15)}{2}$

$a = \frac{b}{2} + \frac{7}{2}$

$b = \frac{16}{- 16}$

$c = - \frac{5 (7 b + 15)}{2}$

$a = \frac{b}{2} + \frac{7}{2}$

Step 3.5.3.3

Simplify the right side.

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Step 3.5.3.3.1

Divide $16$ by $- 16$ .

$b = - 1$

$c = - \frac{5 (7 b + 15)}{2}$

$a = \frac{b}{2} + \frac{7}{2}$

$b = - 1$

$c = - \frac{5 (7 b + 15)}{2}$

$a = \frac{b}{2} + \frac{7}{2}$

$b = - 1$

$c = - \frac{5 (7 b + 15)}{2}$

$a = \frac{b}{2} + \frac{7}{2}$

$b = - 1$

$c = - \frac{5 (7 b + 15)}{2}$

$a = \frac{b}{2} + \frac{7}{2}$

Step 3.6

Replace all occurrences of $b$ with $- 1$ in each equation.

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Step 3.6.1

Replace all occurrences of $b$ in $c = - \frac{5 (7 b + 15)}{2}$ with $- 1$ .

$c = - \frac{5 (7 (- 1) + 15)}{2}$

$b = - 1$

$a = \frac{b}{2} + \frac{7}{2}$

Step 3.6.2

Simplify the right side.

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Step 3.6.2.1

Simplify $- \frac{5 (7 (- 1) + 15)}{2}$ .

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Step 3.6.2.1.1

Simplify the numerator.

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Step 3.6.2.1.1.1

Multiply $7$ by $- 1$ .

$c = - \frac{5 (- 7 + 15)}{2}$

$b = - 1$

$a = \frac{b}{2} + \frac{7}{2}$

Step 3.6.2.1.1.2

Add $- 7$ and $15$ .

$c = - \frac{5 \cdot 8}{2}$

$b = - 1$

$a = \frac{b}{2} + \frac{7}{2}$

$c = - \frac{5 \cdot 8}{2}$

$b = - 1$

$a = \frac{b}{2} + \frac{7}{2}$

Step 3.6.2.1.2

Simplify the expression.

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Step 3.6.2.1.2.1

Multiply $5$ by $8$ .

$c = - \frac{40}{2}$

$b = - 1$

$a = \frac{b}{2} + \frac{7}{2}$

Step 3.6.2.1.2.2

Divide $40$ by $2$ .

$c = - 1 \cdot 20$

$b = - 1$

$a = \frac{b}{2} + \frac{7}{2}$

Step 3.6.2.1.2.3

Multiply $- 1$ by $20$ .

$c = - 20$

$b = - 1$

$a = \frac{b}{2} + \frac{7}{2}$

$c = - 20$

$b = - 1$

$a = \frac{b}{2} + \frac{7}{2}$

$c = - 20$

$b = - 1$

$a = \frac{b}{2} + \frac{7}{2}$

$c = - 20$

$b = - 1$

$a = \frac{b}{2} + \frac{7}{2}$

Step 3.6.3

Replace all occurrences of $b$ in $a = \frac{b}{2} + \frac{7}{2}$ with $- 1$ .

$a = \frac{- 1}{2} + \frac{7}{2}$

$c = - 20$

$b = - 1$

Step 3.6.4

Simplify the right side.

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Step 3.6.4.1

Simplify $\frac{- 1}{2} + \frac{7}{2}$ .

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Step 3.6.4.1.1

Combine the numerators over the common denominator.

$a = \frac{- 1 + 7}{2}$

$c = - 20$

$b = - 1$

Step 3.6.4.1.2

Simplify the expression.

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Step 3.6.4.1.2.1

Add $- 1$ and $7$ .

$a = \frac{6}{2}$

$c = - 20$

$b = - 1$

Step 3.6.4.1.2.2

Divide $6$ by $2$ .

$a = 3$

$c = - 20$

$b = - 1$

$a = 3$

$c = - 20$

$b = - 1$

$a = 3$

$c = - 20$

$b = - 1$

$a = 3$

$c = - 20$

$b = - 1$

$a = 3$

$c = - 20$

$b = - 1$

Step 3.7

List all of the solutions.

$a = 3, c = - 20, b = - 1$

Step 4

Substitute the actual values of $a$ , $b$ , and $c$ into the formula for a quadratic equation to find the resulting equation.

$y = 3 x^{2} - x - 20$

Step 5