Solve for x -f(x)=3x^2+2x-2

$- f (x) = 3 x^{2} + 2 x - 2$

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

$3 x^{2} + 2 x - 2 = - f (x)$

Add $f (x)$ to both sides of the equation.

$3 x^{2} + 2 x - 2 + f (x) = 0$

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 3$ , $b = 2$ , and $c = - 2 + f (x)$ into the quadratic formula and solve for $x$ .

$\frac{- 2 \pm \sqrt{2^{2} - 4 \cdot (3 \cdot (- 2 + f (x)))}}{2 \cdot 3}$

Simplify.

Simplify the numerator.

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

$x = \frac{- 2 \pm \sqrt{4 - 4 \cdot 3 \cdot (- 2 + f (x))}}{2 \cdot 3}$

Multiply $- 4$ by $3$ .

$x = \frac{- 2 \pm \sqrt{4 - 12 \cdot (- 2 + f (x))}}{2 \cdot 3}$

Apply the distributive property.

$x = \frac{- 2 \pm \sqrt{4 - 12 \cdot - 2 - 12 f x}}{2 \cdot 3}$

Multiply $- 12$ by $- 2$ .

$x = \frac{- 2 \pm \sqrt{4 + 24 - 12 f x}}{2 \cdot 3}$

Add $4$ and $24$ .

$x = \frac{- 2 \pm \sqrt{28 - 12 f x}}{2 \cdot 3}$

Rewrite $28 - 12 f (x)$ as $2^{2} (7 - 3 f (x))$ .

Factor $4$ out of $28$ .

$x = \frac{- 2 \pm \sqrt{4 (7) - 12 f x}}{2 \cdot 3}$

Factor $4$ out of $- 12 f (x)$ .

$x = \frac{- 2 \pm \sqrt{4 (7) + 4 (- 3 f x)}}{2 \cdot 3}$

Factor $4$ out of $4 (7) + 4 (- 3 f (x))$ .

$x = \frac{- 2 \pm \sqrt{4 (7 - 3 f x)}}{2 \cdot 3}$

Rewrite $4$ as $2^{2}$ .

$x = \frac{- 2 \pm \sqrt{2^{2} (7 - 3 f x)}}{2 \cdot 3}$

Pull terms out from under the radical.

$x = \frac{- 2 \pm 2 \sqrt{7 - 3 f x}}{2 \cdot 3}$

Multiply $2$ by $3$ .

$x = \frac{- 2 \pm 2 \sqrt{7 - 3 f x}}{6}$

Simplify $\frac{- 2 \pm 2 \sqrt{7 - 3 f (x)}}{6}$ .

$x = \frac{- 1 \pm \sqrt{7 - 3 f x}}{3}$

The final answer is the combination of both solutions.

$x = - \frac{1 - \sqrt{7 - 3 f x}}{3}, - \frac{1 + \sqrt{7 - 3 f x}}{3}$