Algebra Examples

$F (x) = a x^{2} + b x + c$

Step 1

Since $x$ is on the right side of the equation, switch the sides so it is on the left side of the equation.

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

Step 2

Subtract $F x$ from both sides of the equation.

$a x^{2} + b x + c - F x = 0$

Step 3

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 4

Substitute the values $a = a$ , $b = b - F$ , and $c = c$ into the quadratic formula and solve for $x$ .

$\frac{- (b - F) \pm \sqrt{{(b - F)}^{2} - 4 \cdot (a \cdot c)}}{2 a}$

Step 5

Simplify the numerator.

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Step 5.1

Apply the distributive property.

$x = \frac{- b + F \pm \sqrt{{(b - F)}^{2} - 4 \cdot a \cdot c}}{2 a}$

Step 5.2

Multiply $- - F$ .

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Step 5.2.1

Multiply $- 1$ by $- 1$ .

$x = \frac{- b + 1 F \pm \sqrt{{(b - F)}^{2} - 4 \cdot a \cdot c}}{2 a}$

Step 5.2.2

Multiply $F$ by $1$ .

$x = \frac{- b + F \pm \sqrt{{(b - F)}^{2} - 4 \cdot a \cdot c}}{2 a}$

Step 5.3

Rewrite ${(b - F)}^{2}$ as $(b - F) (b - F)$ .

$x = \frac{- b + F \pm \sqrt{(b - F) (b - F) - 4 \cdot a \cdot c}}{2 a}$

Step 5.4

Expand $(b - F) (b - F)$ using the FOIL Method.

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Step 5.4.1

Apply the distributive property.

$x = \frac{- b + F \pm \sqrt{b (b - F) - F (b - F) - 4 \cdot a \cdot c}}{2 a}$

Step 5.4.2

Apply the distributive property.

$x = \frac{- b + F \pm \sqrt{b \cdot b + b (- F) - F (b - F) - 4 \cdot a \cdot c}}{2 a}$

Step 5.4.3

Apply the distributive property.

$x = \frac{- b + F \pm \sqrt{b \cdot b + b (- F) - F b - F (- F) - 4 \cdot a \cdot c}}{2 a}$

Step 5.5

Simplify and combine like terms.

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Step 5.5.1

Simplify each term.

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Step 5.5.1.1

Multiply $b$ by $b$ .

$x = \frac{- b + F \pm \sqrt{b^{2} + b (- F) - F b - F (- F) - 4 \cdot a \cdot c}}{2 a}$

Step 5.5.1.2

Rewrite using the commutative property of multiplication.

$x = \frac{- b + F \pm \sqrt{b^{2} - b F - F b - F (- F) - 4 \cdot a \cdot c}}{2 a}$

Step 5.5.1.3

Rewrite using the commutative property of multiplication.

$x = \frac{- b + F \pm \sqrt{b^{2} - b F - F b - 1 \cdot (- 1 F \cdot F) - 4 \cdot a \cdot c}}{2 a}$

Step 5.5.1.4

Multiply $F$ by $F$ by adding the exponents.

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Step 5.5.1.4.1

Move $F$ .

$x = \frac{- b + F \pm \sqrt{b^{2} - b F - F b - 1 \cdot (- 1 (F \cdot F)) - 4 \cdot a \cdot c}}{2 a}$

Step 5.5.1.4.2

Multiply $F$ by $F$ .

$x = \frac{- b + F \pm \sqrt{b^{2} - b F - F b - 1 \cdot (- 1 F^{2}) - 4 \cdot a \cdot c}}{2 a}$

Step 5.5.1.5

Multiply $- 1$ by $- 1$ .

$x = \frac{- b + F \pm \sqrt{b^{2} - b F - F b + 1 F^{2} - 4 \cdot a \cdot c}}{2 a}$

Step 5.5.1.6

Multiply $F^{2}$ by $1$ .

$x = \frac{- b + F \pm \sqrt{b^{2} - b F - F b + F^{2} - 4 \cdot a \cdot c}}{2 a}$

Step 5.5.2

Subtract $F b$ from $- b F$ .

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Step 5.5.2.1

Move $F$ .

$x = \frac{- b + F \pm \sqrt{b^{2} - b F - 1 b F + F^{2} - 4 \cdot a \cdot c}}{2 a}$

Step 5.5.2.2

Subtract $b F$ from $- b F$ .

$x = \frac{- b + F \pm \sqrt{b^{2} - 2 b F + F^{2} - 4 \cdot a \cdot c}}{2 a}$

$x = \frac{- b + F \pm \sqrt{b^{2} - 2 b F + F^{2} - 4 a c}}{2 a}$

Step 6

The final answer is the combination of both solutions.

$x = - \frac{b - F - \sqrt{b^{2} - 2 b F + F^{2} - 4 a c}}{2 a}$

$x = - \frac{b - F + \sqrt{b^{2} - 2 b F + F^{2} - 4 a c}}{2 a}$