Solve for x f(-3)=4x^2-x-3

$f (- 3) = 4 x^{2} - x - 3$

Rewrite the equation as $4 x^{2} - x - 3 = f (- 3)$ .

$4 x^{2} - x - 3 = f (- 3)$

Move $- 3$ to the left of $f$ .

$4 x^{2} - x - 3 = - 3 f$

Add $3 f$ to both sides of the equation.

$4 x^{2} - x - 3 + 3 f = 0$

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 4$ , $b = - 1$ , and $c = - 3 + 3 f$ into the quadratic formula and solve for $x$ .

$\frac{1 \pm \sqrt{{(- 1)}^{2} - 4 \cdot (4 \cdot (- 3 + 3 f))}}{2 \cdot 4}$

Simplify.

Simplify the numerator.

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

$x = \frac{1 \pm \sqrt{1 - 4 \cdot 4 \cdot (- 3 + 3 f)}}{2 \cdot 4}$

Multiply $- 4$ by $4$ .

$x = \frac{1 \pm \sqrt{1 - 16 \cdot (- 3 + 3 f)}}{2 \cdot 4}$

Apply the distributive property.

$x = \frac{1 \pm \sqrt{1 - 16 \cdot - 3 - 16 (3 f)}}{2 \cdot 4}$

Multiply $- 16$ by $- 3$ .

$x = \frac{1 \pm \sqrt{1 + 48 - 16 (3 f)}}{2 \cdot 4}$

Multiply $3$ by $- 16$ .

$x = \frac{1 \pm \sqrt{1 + 48 - 48 f}}{2 \cdot 4}$

Add $1$ and $48$ .

$x = \frac{1 \pm \sqrt{49 - 48 f}}{2 \cdot 4}$

Multiply $2$ by $4$ .

$x = \frac{1 \pm \sqrt{49 - 48 f}}{8}$

The final answer is the combination of both solutions.

$x = \frac{1 + \sqrt{49 - 48 f}}{8}$

$x = \frac{1 - \sqrt{49 - 48 f}}{8}$