Algebra Examples

$(- 2, - 16)$ , $(3, 11)$ , $(0, 2)$

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.

$- 16 = a {(- 2)}^{2} + b (- 2) + c, 11 = a {(3)}^{2} + b (3) + c, 2 = a {(0)}^{2} + b (0) + c$

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 $2 = a \cdot 0^{2} + c$ .

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Step 3.1.1

Rewrite the equation as $a \cdot 0^{2} + c = 2$ .

$a \cdot 0^{2} + c = 2$

$- 16 = a {(- 2)}^{2} + b (- 2) + c$

$11 = a \cdot 3^{2} + b (3) + c$

Step 3.1.2

Simplify $a \cdot 0^{2} + c$ .

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Step 3.1.2.1

Simplify each term.

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Step 3.1.2.1.1

Raising $0$ to any positive power yields $0$ .

$a \cdot 0 + c = 2$

$- 16 = a {(- 2)}^{2} + b (- 2) + c$

$11 = a \cdot 3^{2} + b (3) + c$

Step 3.1.2.1.2

Multiply $a$ by $0$ .

$0 + c = 2$

$- 16 = a {(- 2)}^{2} + b (- 2) + c$

$11 = a \cdot 3^{2} + b (3) + c$

$0 + c = 2$

$- 16 = a {(- 2)}^{2} + b (- 2) + c$

$11 = a \cdot 3^{2} + b (3) + c$

Step 3.1.2.2

Add $0$ and $c$ .

$c = 2$

$- 16 = a {(- 2)}^{2} + b (- 2) + c$

$11 = a \cdot 3^{2} + b (3) + c$

$c = 2$

$- 16 = a {(- 2)}^{2} + b (- 2) + c$

$11 = a \cdot 3^{2} + b (3) + c$

$c = 2$

$- 16 = a {(- 2)}^{2} + b (- 2) + c$

$11 = a \cdot 3^{2} + b (3) + c$

Step 3.2

Replace all occurrences of $c$ with $2$ in each equation.

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Step 3.2.1

Replace all occurrences of $c$ in $- 16 = a {(- 2)}^{2} + b (- 2) + c$ with $2$ .

$- 16 = a {(- 2)}^{2} + b (- 2) + 2$

$c = 2$

$11 = a \cdot 3^{2} + b (3) + c$

Step 3.2.2

Simplify $- 16 = a {(- 2)}^{2} + b (- 2) + 2$ .

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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.

$- 16 = a {(- 2)}^{2} + b (- 2) + 2$

$c = 2$

$11 = a \cdot 3^{2} + b (3) + c$

$- 16 = a {(- 2)}^{2} + b (- 2) + 2$

$c = 2$

$11 = a \cdot 3^{2} + b (3) + c$

Step 3.2.2.2

Simplify the right side.

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Step 3.2.2.2.1

Simplify each term.

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Step 3.2.2.2.1.1

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

$- 16 = a \cdot 4 + b (- 2) + 2$

$c = 2$

$11 = a \cdot 3^{2} + b (3) + c$

Step 3.2.2.2.1.2

Move $4$ to the left of $a$ .

$- 16 = 4 \cdot a + b (- 2) + 2$

$c = 2$

$11 = a \cdot 3^{2} + b (3) + c$

Step 3.2.2.2.1.3

Move $- 2$ to the left of $b$ .

$- 16 = 4 a - 2 b + 2$

$c = 2$

$11 = a \cdot 3^{2} + b (3) + c$

$- 16 = 4 a - 2 b + 2$

$c = 2$

$11 = a \cdot 3^{2} + b (3) + c$

$- 16 = 4 a - 2 b + 2$

$c = 2$

$11 = a \cdot 3^{2} + b (3) + c$

$- 16 = 4 a - 2 b + 2$

$c = 2$

$11 = a \cdot 3^{2} + b (3) + c$

Step 3.2.3

Replace all occurrences of $c$ in $11 = a \cdot 3^{2} + b (3) + c$ with $2$ .

$11 = a \cdot 3^{2} + b (3) + 2$

$- 16 = 4 a - 2 b + 2$

$c = 2$

Step 3.2.4

Simplify $11 = a \cdot 3^{2} + b (3) + 2$ .

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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.

$11 = a \cdot 3^{2} + b (3) + 2$

$- 16 = 4 a - 2 b + 2$

$c = 2$

$11 = a \cdot 3^{2} + b (3) + 2$

$- 16 = 4 a - 2 b + 2$

$c = 2$

Step 3.2.4.2

Simplify the right side.

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Step 3.2.4.2.1

Simplify each term.

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Step 3.2.4.2.1.1

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

$11 = a \cdot 9 + b (3) + 2$

$- 16 = 4 a - 2 b + 2$

$c = 2$

Step 3.2.4.2.1.2

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

$11 = 9 \cdot a + b (3) + 2$

$- 16 = 4 a - 2 b + 2$

$c = 2$

Step 3.2.4.2.1.3

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

$11 = 9 a + 3 b + 2$

$- 16 = 4 a - 2 b + 2$

$c = 2$

$11 = 9 a + 3 b + 2$

$- 16 = 4 a - 2 b + 2$

$c = 2$

$11 = 9 a + 3 b + 2$

$- 16 = 4 a - 2 b + 2$

$c = 2$

$11 = 9 a + 3 b + 2$

$- 16 = 4 a - 2 b + 2$

$c = 2$

$11 = 9 a + 3 b + 2$

$- 16 = 4 a - 2 b + 2$

$c = 2$

Step 3.3

Solve for $a$ in $11 = 9 a + 3 b + 2$ .

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Step 3.3.1

Rewrite the equation as $9 a + 3 b + 2 = 11$ .

$9 a + 3 b + 2 = 11$

$- 16 = 4 a - 2 b + 2$

$c = 2$

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 $3 b$ from both sides of the equation.

$9 a + 2 = 11 - 3 b$

$- 16 = 4 a - 2 b + 2$

$c = 2$

Step 3.3.2.2

Subtract $2$ from both sides of the equation.

$9 a = 11 - 3 b - 2$

$- 16 = 4 a - 2 b + 2$

$c = 2$

Step 3.3.2.3

Subtract $2$ from $11$ .

$9 a = - 3 b + 9$

$- 16 = 4 a - 2 b + 2$

$c = 2$

$9 a = - 3 b + 9$

$- 16 = 4 a - 2 b + 2$

$c = 2$

Step 3.3.3

Divide each term in $9 a = - 3 b + 9$ by $9$ and simplify.

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Step 3.3.3.1

Divide each term in $9 a = - 3 b + 9$ by $9$ .

$\frac{9 a}{9} = \frac{- 3 b}{9} + \frac{9}{9}$

$- 16 = 4 a - 2 b + 2$

$c = 2$

Step 3.3.3.2

Simplify the left side.

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Step 3.3.3.2.1

Cancel the common factor of $9$ .

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Step 3.3.3.2.1.1

Cancel the common factor.

$\frac{9 a}{9} = \frac{- 3 b}{9} + \frac{9}{9}$

$- 16 = 4 a - 2 b + 2$

$c = 2$

Step 3.3.3.2.1.2

Divide $a$ by $1$ .

$a = \frac{- 3 b}{9} + \frac{9}{9}$

$- 16 = 4 a - 2 b + 2$

$c = 2$

$a = \frac{- 3 b}{9} + \frac{9}{9}$

$- 16 = 4 a - 2 b + 2$

$c = 2$

$a = \frac{- 3 b}{9} + \frac{9}{9}$

$- 16 = 4 a - 2 b + 2$

$c = 2$

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 $- 3$ and $9$ .

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Step 3.3.3.3.1.1.1

Factor $3$ out of $- 3 b$ .

$a = \frac{3 (- b)}{9} + \frac{9}{9}$

$- 16 = 4 a - 2 b + 2$

$c = 2$

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 $3$ out of $9$ .

$a = \frac{3 (- b)}{3 (3)} + \frac{9}{9}$

$- 16 = 4 a - 2 b + 2$

$c = 2$

Step 3.3.3.3.1.1.2.2

Cancel the common factor.

$a = \frac{3 (- b)}{3 \cdot 3} + \frac{9}{9}$

$- 16 = 4 a - 2 b + 2$

$c = 2$

Step 3.3.3.3.1.1.2.3

Rewrite the expression.

$a = \frac{- b}{3} + \frac{9}{9}$

$- 16 = 4 a - 2 b + 2$

$c = 2$

$a = \frac{- b}{3} + \frac{9}{9}$

$- 16 = 4 a - 2 b + 2$

$c = 2$

$a = \frac{- b}{3} + \frac{9}{9}$

$- 16 = 4 a - 2 b + 2$

$c = 2$

Step 3.3.3.3.1.2

Move the negative in front of the fraction.

$a = - \frac{b}{3} + \frac{9}{9}$

$- 16 = 4 a - 2 b + 2$

$c = 2$

Step 3.3.3.3.1.3

Divide $9$ by $9$ .

$a = - \frac{b}{3} + 1$

$- 16 = 4 a - 2 b + 2$

$c = 2$

$a = - \frac{b}{3} + 1$

$- 16 = 4 a - 2 b + 2$

$c = 2$

$a = - \frac{b}{3} + 1$

$- 16 = 4 a - 2 b + 2$

$c = 2$

$a = - \frac{b}{3} + 1$

$- 16 = 4 a - 2 b + 2$

$c = 2$

$a = - \frac{b}{3} + 1$

$- 16 = 4 a - 2 b + 2$

$c = 2$

Step 3.4

Replace all occurrences of $a$ with $- \frac{b}{3} + 1$ in each equation.

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Step 3.4.1

Replace all occurrences of $a$ in $- 16 = 4 a - 2 b + 2$ with $- \frac{b}{3} + 1$ .

$- 16 = 4 (- \frac{b}{3} + 1) - 2 b + 2$

$a = - \frac{b}{3} + 1$

$c = 2$

Step 3.4.2

Simplify the right side.

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Step 3.4.2.1

Simplify $4 (- \frac{b}{3} + 1) - 2 b + 2$ .

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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.

$- 16 = 4 (- \frac{b}{3}) + 4 \cdot 1 - 2 b + 2$

$a = - \frac{b}{3} + 1$

$c = 2$

Step 3.4.2.1.1.2

Multiply $4 (- \frac{b}{3})$ .

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Step 3.4.2.1.1.2.1

Multiply $- 1$ by $4$ .

$- 16 = - 4 \frac{b}{3} + 4 \cdot 1 - 2 b + 2$

$a = - \frac{b}{3} + 1$

$c = 2$

Step 3.4.2.1.1.2.2

Combine $- 4$ and $\frac{b}{3}$ .

$- 16 = \frac{- 4 b}{3} + 4 \cdot 1 - 2 b + 2$

$a = - \frac{b}{3} + 1$

$c = 2$

$- 16 = \frac{- 4 b}{3} + 4 \cdot 1 - 2 b + 2$

$a = - \frac{b}{3} + 1$

$c = 2$

Step 3.4.2.1.1.3

Multiply $4$ by $1$ .

$- 16 = \frac{- 4 b}{3} + 4 - 2 b + 2$

$a = - \frac{b}{3} + 1$

$c = 2$

Step 3.4.2.1.1.4

Move the negative in front of the fraction.

$- 16 = - \frac{4 b}{3} + 4 - 2 b + 2$

$a = - \frac{b}{3} + 1$

$c = 2$

$- 16 = - \frac{4 b}{3} + 4 - 2 b + 2$

$a = - \frac{b}{3} + 1$

$c = 2$

Step 3.4.2.1.2

To write $- 2 b$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$- 16 = - \frac{4 b}{3} - 2 b \cdot \frac{3}{3} + 4 + 2$

$a = - \frac{b}{3} + 1$

$c = 2$

Step 3.4.2.1.3

Combine $- 2 b$ and $\frac{3}{3}$ .

$- 16 = - \frac{4 b}{3} + \frac{- 2 b \cdot 3}{3} + 4 + 2$

$a = - \frac{b}{3} + 1$

$c = 2$

Step 3.4.2.1.4

Combine the numerators over the common denominator.

$- 16 = \frac{- 4 b - 2 b \cdot 3}{3} + 4 + 2$

$a = - \frac{b}{3} + 1$

$c = 2$

Step 3.4.2.1.5

Find the common denominator.

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Step 3.4.2.1.5.1

Write $4$ as a fraction with denominator $1$ .

$- 16 = \frac{- 4 b - 2 b \cdot 3}{3} + \frac{4}{1} + 2$

$a = - \frac{b}{3} + 1$

$c = 2$

Step 3.4.2.1.5.2

Multiply $\frac{4}{1}$ by $\frac{3}{3}$ .

$- 16 = \frac{- 4 b - 2 b \cdot 3}{3} + \frac{4}{1} \cdot \frac{3}{3} + 2$

$a = - \frac{b}{3} + 1$

$c = 2$

Step 3.4.2.1.5.3

Multiply $\frac{4}{1}$ by $\frac{3}{3}$ .

$- 16 = \frac{- 4 b - 2 b \cdot 3}{3} + \frac{4 \cdot 3}{3} + 2$

$a = - \frac{b}{3} + 1$

$c = 2$

Step 3.4.2.1.5.4

Write $2$ as a fraction with denominator $1$ .

$- 16 = \frac{- 4 b - 2 b \cdot 3}{3} + \frac{4 \cdot 3}{3} + \frac{2}{1}$

$a = - \frac{b}{3} + 1$

$c = 2$

Step 3.4.2.1.5.5

Multiply $\frac{2}{1}$ by $\frac{3}{3}$ .

$- 16 = \frac{- 4 b - 2 b \cdot 3}{3} + \frac{4 \cdot 3}{3} + \frac{2}{1} \cdot \frac{3}{3}$

$a = - \frac{b}{3} + 1$

$c = 2$

Step 3.4.2.1.5.6

Multiply $\frac{2}{1}$ by $\frac{3}{3}$ .

$- 16 = \frac{- 4 b - 2 b \cdot 3}{3} + \frac{4 \cdot 3}{3} + \frac{2 \cdot 3}{3}$

$a = - \frac{b}{3} + 1$

$c = 2$

$- 16 = \frac{- 4 b - 2 b \cdot 3}{3} + \frac{4 \cdot 3}{3} + \frac{2 \cdot 3}{3}$

$a = - \frac{b}{3} + 1$

$c = 2$

Step 3.4.2.1.6

Combine the numerators over the common denominator.

$- 16 = \frac{- 4 b - 2 b \cdot 3 + 4 \cdot 3 + 2 \cdot 3}{3}$

$a = - \frac{b}{3} + 1$

$c = 2$

Step 3.4.2.1.7

Simplify each term.

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Step 3.4.2.1.7.1

Multiply $3$ by $- 2$ .

$- 16 = \frac{- 4 b - 6 b + 4 \cdot 3 + 2 \cdot 3}{3}$

$a = - \frac{b}{3} + 1$

$c = 2$

Step 3.4.2.1.7.2

Multiply $4$ by $3$ .

$- 16 = \frac{- 4 b - 6 b + 12 + 2 \cdot 3}{3}$

$a = - \frac{b}{3} + 1$

$c = 2$

Step 3.4.2.1.7.3

Multiply $2$ by $3$ .

$- 16 = \frac{- 4 b - 6 b + 12 + 6}{3}$

$a = - \frac{b}{3} + 1$

$c = 2$

$- 16 = \frac{- 4 b - 6 b + 12 + 6}{3}$

$a = - \frac{b}{3} + 1$

$c = 2$

Step 3.4.2.1.8

Simplify terms.

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Step 3.4.2.1.8.1

Subtract $6 b$ from $- 4 b$ .

$- 16 = \frac{- 10 b + 12 + 6}{3}$

$a = - \frac{b}{3} + 1$

$c = 2$

Step 3.4.2.1.8.2

Add $12$ and $6$ .

$- 16 = \frac{- 10 b + 18}{3}$

$a = - \frac{b}{3} + 1$

$c = 2$

Step 3.4.2.1.8.3

Factor $2$ out of $- 10 b + 18$ .

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Step 3.4.2.1.8.3.1

Factor $2$ out of $- 10 b$ .

$- 16 = \frac{2 (- 5 b) + 18}{3}$

$a = - \frac{b}{3} + 1$

$c = 2$

Step 3.4.2.1.8.3.2

Factor $2$ out of $18$ .

$- 16 = \frac{2 (- 5 b) + 2 (9)}{3}$

$a = - \frac{b}{3} + 1$

$c = 2$

Step 3.4.2.1.8.3.3

Factor $2$ out of $2 (- 5 b) + 2 (9)$ .

$- 16 = \frac{2 (- 5 b + 9)}{3}$

$a = - \frac{b}{3} + 1$

$c = 2$

$- 16 = \frac{2 (- 5 b + 9)}{3}$

$a = - \frac{b}{3} + 1$

$c = 2$

Step 3.4.2.1.8.4

Factor $- 1$ out of $- 5 b$ .

$- 16 = \frac{2 (- (5 b) + 9)}{3}$

$a = - \frac{b}{3} + 1$

$c = 2$

Step 3.4.2.1.8.5

Rewrite $9$ as $- 1 (- 9)$ .

$- 16 = \frac{2 (- (5 b) - 1 \cdot - 9)}{3}$

$a = - \frac{b}{3} + 1$

$c = 2$

Step 3.4.2.1.8.6

Factor $- 1$ out of $- (5 b) - 1 (- 9)$ .

$- 16 = \frac{2 (- (5 b - 9))}{3}$

$a = - \frac{b}{3} + 1$

$c = 2$

Step 3.4.2.1.8.7

Simplify the expression.

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Step 3.4.2.1.8.7.1

Rewrite $- (5 b - 9)$ as $- 1 (5 b - 9)$ .

$- 16 = \frac{2 (- 1 (5 b - 9))}{3}$

$a = - \frac{b}{3} + 1$

$c = 2$

Step 3.4.2.1.8.7.2

Move the negative in front of the fraction.

$- 16 = - \frac{2 (5 b - 9)}{3}$

$a = - \frac{b}{3} + 1$

$c = 2$

$- 16 = - \frac{2 (5 b - 9)}{3}$

$a = - \frac{b}{3} + 1$

$c = 2$

$- 16 = - \frac{2 (5 b - 9)}{3}$

$a = - \frac{b}{3} + 1$

$c = 2$

$- 16 = - \frac{2 (5 b - 9)}{3}$

$a = - \frac{b}{3} + 1$

$c = 2$

$- 16 = - \frac{2 (5 b - 9)}{3}$

$a = - \frac{b}{3} + 1$

$c = 2$

$- 16 = - \frac{2 (5 b - 9)}{3}$

$a = - \frac{b}{3} + 1$

$c = 2$

Step 3.5

Solve for $b$ in $- 16 = - \frac{2 (5 b - 9)}{3}$ .

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Step 3.5.1

Rewrite the equation as $- \frac{2 (5 b - 9)}{3} = - 16$ .

$- \frac{2 (5 b - 9)}{3} = - 16$

$a = - \frac{b}{3} + 1$

$c = 2$

Step 3.5.2

Multiply both sides of the equation by $- \frac{3}{2}$ .

$- \frac{3}{2} \cdot (- \frac{2 (5 b - 9)}{3}) = - \frac{3}{2} \cdot - 16$

$a = - \frac{b}{3} + 1$

$c = 2$

Step 3.5.3

Simplify both sides of the equation.

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Step 3.5.3.1

Simplify the left side.

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Step 3.5.3.1.1

Simplify $- \frac{3}{2} (- \frac{2 (5 b - 9)}{3})$ .

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Step 3.5.3.1.1.1

Cancel the common factor of $3$ .

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Step 3.5.3.1.1.1.1

Move the leading negative in $- \frac{3}{2}$ into the numerator.

$\frac{- 3}{2} \cdot (- \frac{2 (5 b - 9)}{3}) = - \frac{3}{2} \cdot - 16$

$a = - \frac{b}{3} + 1$

$c = 2$

Step 3.5.3.1.1.1.2

Move the leading negative in $- \frac{2 (5 b - 9)}{3}$ into the numerator.

$\frac{- 3}{2} \cdot \frac{- 2 (5 b - 9)}{3} = - \frac{3}{2} \cdot - 16$

$a = - \frac{b}{3} + 1$

$c = 2$

Step 3.5.3.1.1.1.3

Factor $3$ out of $- 3$ .

$\frac{3 (- 1)}{2} \cdot \frac{- 2 (5 b - 9)}{3} = - \frac{3}{2} \cdot - 16$

$a = - \frac{b}{3} + 1$

$c = 2$

Step 3.5.3.1.1.1.4

Cancel the common factor.

$\frac{3 \cdot - 1}{2} \cdot \frac{- 2 (5 b - 9)}{3} = - \frac{3}{2} \cdot - 16$

$a = - \frac{b}{3} + 1$

$c = 2$

Step 3.5.3.1.1.1.5

Rewrite the expression.

$\frac{- 1}{2} \cdot (- 2 (5 b - 9)) = - \frac{3}{2} \cdot - 16$

$a = - \frac{b}{3} + 1$

$c = 2$

$\frac{- 1}{2} \cdot (- 2 (5 b - 9)) = - \frac{3}{2} \cdot - 16$

$a = - \frac{b}{3} + 1$

$c = 2$

Step 3.5.3.1.1.2

Cancel the common factor of $2$ .

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Step 3.5.3.1.1.2.1

Factor $2$ out of $- 2 (5 b - 9)$ .

$\frac{- 1}{2} \cdot (2 (- (5 b - 9))) = - \frac{3}{2} \cdot - 16$

$a = - \frac{b}{3} + 1$

$c = 2$

Step 3.5.3.1.1.2.2

Cancel the common factor.

$\frac{- 1}{2} \cdot (2 (- (5 b - 9))) = - \frac{3}{2} \cdot - 16$

$a = - \frac{b}{3} + 1$

$c = 2$

Step 3.5.3.1.1.2.3

Rewrite the expression.

$5 b - 9 = - \frac{3}{2} \cdot - 16$

$a = - \frac{b}{3} + 1$

$c = 2$

$5 b - 9 = - \frac{3}{2} \cdot - 16$

$a = - \frac{b}{3} + 1$

$c = 2$

Step 3.5.3.1.1.3

Multiply.

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Step 3.5.3.1.1.3.1

Multiply $- 1$ by $- 1$ .

$1 (5 b - 9) = - \frac{3}{2} \cdot - 16$

$a = - \frac{b}{3} + 1$

$c = 2$

Step 3.5.3.1.1.3.2

Multiply $5 b - 9$ by $1$ .

$5 b - 9 = - \frac{3}{2} \cdot - 16$

$a = - \frac{b}{3} + 1$

$c = 2$

$5 b - 9 = - \frac{3}{2} \cdot - 16$

$a = - \frac{b}{3} + 1$

$c = 2$

$5 b - 9 = - \frac{3}{2} \cdot - 16$

$a = - \frac{b}{3} + 1$

$c = 2$

$5 b - 9 = - \frac{3}{2} \cdot - 16$

$a = - \frac{b}{3} + 1$

$c = 2$

Step 3.5.3.2

Simplify the right side.

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Step 3.5.3.2.1

Simplify $- \frac{3}{2} \cdot - 16$ .

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Step 3.5.3.2.1.1

Cancel the common factor of $2$ .

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Step 3.5.3.2.1.1.1

Move the leading negative in $- \frac{3}{2}$ into the numerator.

$5 b - 9 = \frac{- 3}{2} \cdot - 16$

$a = - \frac{b}{3} + 1$

$c = 2$

Step 3.5.3.2.1.1.2

Factor $2$ out of $- 16$ .

$5 b - 9 = \frac{- 3}{2} \cdot (2 (- 8))$

$a = - \frac{b}{3} + 1$

$c = 2$

Step 3.5.3.2.1.1.3

Cancel the common factor.

$5 b - 9 = \frac{- 3}{2} \cdot (2 \cdot - 8)$

$a = - \frac{b}{3} + 1$

$c = 2$

Step 3.5.3.2.1.1.4

Rewrite the expression.

$5 b - 9 = - 3 \cdot - 8$

$a = - \frac{b}{3} + 1$

$c = 2$

$5 b - 9 = - 3 \cdot - 8$

$a = - \frac{b}{3} + 1$

$c = 2$

Step 3.5.3.2.1.2

Multiply $- 3$ by $- 8$ .

$5 b - 9 = 24$

$a = - \frac{b}{3} + 1$

$c = 2$

$5 b - 9 = 24$

$a = - \frac{b}{3} + 1$

$c = 2$

$5 b - 9 = 24$

$a = - \frac{b}{3} + 1$

$c = 2$

$5 b - 9 = 24$

$a = - \frac{b}{3} + 1$

$c = 2$

Step 3.5.4

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

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Step 3.5.4.1

Add $9$ to both sides of the equation.

$5 b = 24 + 9$

$a = - \frac{b}{3} + 1$

$c = 2$

Step 3.5.4.2

Add $24$ and $9$ .

$5 b = 33$

$a = - \frac{b}{3} + 1$

$c = 2$

$5 b = 33$

$a = - \frac{b}{3} + 1$

$c = 2$

Step 3.5.5

Divide each term in $5 b = 33$ by $5$ and simplify.

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Step 3.5.5.1

Divide each term in $5 b = 33$ by $5$ .

$\frac{5 b}{5} = \frac{33}{5}$

$a = - \frac{b}{3} + 1$

$c = 2$

Step 3.5.5.2

Simplify the left side.

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Step 3.5.5.2.1

Cancel the common factor of $5$ .

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Step 3.5.5.2.1.1

Cancel the common factor.

$\frac{5 b}{5} = \frac{33}{5}$

$a = - \frac{b}{3} + 1$

$c = 2$

Step 3.5.5.2.1.2

Divide $b$ by $1$ .

$b = \frac{33}{5}$

$a = - \frac{b}{3} + 1$

$c = 2$

$b = \frac{33}{5}$

$a = - \frac{b}{3} + 1$

$c = 2$

$b = \frac{33}{5}$

$a = - \frac{b}{3} + 1$

$c = 2$

$b = \frac{33}{5}$

$a = - \frac{b}{3} + 1$

$c = 2$

$b = \frac{33}{5}$

$a = - \frac{b}{3} + 1$

$c = 2$

Step 3.6

Replace all occurrences of $b$ with $\frac{33}{5}$ in each equation.

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Step 3.6.1

Replace all occurrences of $b$ in $a = - \frac{b}{3} + 1$ with $\frac{33}{5}$ .

$a = - \frac{\frac{33}{5}}{3} + 1$

$b = \frac{33}{5}$

$c = 2$

Step 3.6.2

Simplify the right side.

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Step 3.6.2.1

Simplify $- \frac{\frac{33}{5}}{3} + 1$ .

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Step 3.6.2.1.1

Simplify each term.

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Step 3.6.2.1.1.1

Multiply the numerator by the reciprocal of the denominator.

$a = - (\frac{33}{5} \cdot \frac{1}{3}) + 1$

$b = \frac{33}{5}$

$c = 2$

Step 3.6.2.1.1.2

Cancel the common factor of $3$ .

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Step 3.6.2.1.1.2.1

Factor $3$ out of $33$ .

$a = - (\frac{3 (11)}{5} \cdot \frac{1}{3}) + 1$

$b = \frac{33}{5}$

$c = 2$

Step 3.6.2.1.1.2.2

Cancel the common factor.

$a = - (\frac{3 \cdot 11}{5} \cdot \frac{1}{3}) + 1$

$b = \frac{33}{5}$

$c = 2$

Step 3.6.2.1.1.2.3

Rewrite the expression.

$a = - \frac{11}{5} + 1$

$b = \frac{33}{5}$

$c = 2$

$a = - \frac{11}{5} + 1$

$b = \frac{33}{5}$

$c = 2$

$a = - \frac{11}{5} + 1$

$b = \frac{33}{5}$

$c = 2$

Step 3.6.2.1.2

Simplify the expression.

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Step 3.6.2.1.2.1

Write $1$ as a fraction with a common denominator.

$a = - \frac{11}{5} + \frac{5}{5}$

$b = \frac{33}{5}$

$c = 2$

Step 3.6.2.1.2.2

Combine the numerators over the common denominator.

$a = \frac{- 11 + 5}{5}$

$b = \frac{33}{5}$

$c = 2$

Step 3.6.2.1.2.3

Add $- 11$ and $5$ .

$a = \frac{- 6}{5}$

$b = \frac{33}{5}$

$c = 2$

Step 3.6.2.1.2.4

Move the negative in front of the fraction.

$a = - \frac{6}{5}$

$b = \frac{33}{5}$

$c = 2$

$a = - \frac{6}{5}$

$b = \frac{33}{5}$

$c = 2$

$a = - \frac{6}{5}$

$b = \frac{33}{5}$

$c = 2$

$a = - \frac{6}{5}$

$b = \frac{33}{5}$

$c = 2$

$a = - \frac{6}{5}$

$b = \frac{33}{5}$

$c = 2$

Step 3.7

List all of the solutions.

$a = - \frac{6}{5}, b = \frac{33}{5}, c = 2$

Step 4

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

$y = - \frac{6 x^{2}}{5} + \frac{33 x}{5} + 2$

Step 5