Solve for x f(-1)=2x^2-x-1

$f (- 1) = 2 x^{2} - x - 1$

Rewrite the equation as $2 x^{2} - x - 1 = f (- 1)$ .

$2 x^{2} - x - 1 = f (- 1)$

Simplify $f (- 1)$ .

Move $- 1$ to the left of $f$ .

$2 x^{2} - x - 1 = - 1 f$

Rewrite $- 1 f$ as $- f$ .

$2 x^{2} - x - 1 = - f$

Add $f$ to both sides of the equation.

$2 x^{2} - x - 1 + f = 0$

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 2$ , $b = - 1$ , and $c = - 1 + f$ into the quadratic formula and solve for $x$ .

$\frac{1 \pm \sqrt{{(- 1)}^{2} - 4 \cdot (2 \cdot (- 1 + f))}}{2 \cdot 2}$

Simplify.

Simplify the numerator.

Raise $- 1$ to the power of $2$ .

$x = \frac{1 \pm \sqrt{1 - 4 \cdot 2 \cdot (- 1 + f)}}{2 \cdot 2}$

Multiply $- 4$ by $2$ .

$x = \frac{1 \pm \sqrt{1 - 8 \cdot (- 1 + f)}}{2 \cdot 2}$

Apply the distributive property.

$x = \frac{1 \pm \sqrt{1 - 8 \cdot - 1 - 8 f}}{2 \cdot 2}$

Multiply $- 8$ by $- 1$ .

$x = \frac{1 \pm \sqrt{1 + 8 - 8 f}}{2 \cdot 2}$

Add $1$ and $8$ .

$x = \frac{1 \pm \sqrt{9 - 8 f}}{2 \cdot 2}$

Multiply $2$ by $2$ .

$x = \frac{1 \pm \sqrt{9 - 8 f}}{4}$

The final answer is the combination of both solutions.

$x = \frac{1 + \sqrt{9 - 8 f}}{4}$

$x = \frac{1 - \sqrt{9 - 8 f}}{4}$