Solve for x F(x)=ax^2+bx+c

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

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.

Apply the distributive property.

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

Multiply $- - F$ .

Multiply $- 1$ by $- 1$ .

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

Multiply $F$ by $1$ .

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

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

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

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

Apply the distributive property.

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

Apply the distributive property.

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

Apply the distributive property.

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

Simplify and combine like terms.

Simplify each term.

Multiply $b$ by $b$ .

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

Rewrite using the commutative property of multiplication.

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

Rewrite using the commutative property of multiplication.

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

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

Move $F$ .

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

Multiply $F$ by $F$ .

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

Multiply $- 1$ by $- 1$ .

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

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

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

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

Move $F$ .

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

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.

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

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

Solve for x F(x)=ax^2+bx+c

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