Find the Roots (Zeros) f(x)=x^2+8x-96

$f (x) = x^{2} + 8 x - 96$

Step 1

Set $x^{2} + 8 x - 96$ equal to $0$ .

$x^{2} + 8 x - 96 = 0$

Step 2

Solve for $x$ .

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 1$ , $b = 8$ , and $c = - 96$ into the quadratic formula and solve for $x$ .

$\frac{- 8 \pm \sqrt{8^{2} - 4 \cdot (1 \cdot - 96)}}{2 \cdot 1}$

Simplify.

Simplify the numerator.

Raise $8$ to the power of $2$ .

$x = \frac{- 8 \pm \sqrt{64 - 4 \cdot 1 \cdot - 96}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot - 96$ .

Multiply $- 4$ by $1$ .

$x = \frac{- 8 \pm \sqrt{64 - 4 \cdot - 96}}{2 \cdot 1}$

Multiply $- 4$ by $- 96$ .

$x = \frac{- 8 \pm \sqrt{64 + 384}}{2 \cdot 1}$

Add $64$ and $384$ .

$x = \frac{- 8 \pm \sqrt{448}}{2 \cdot 1}$

Rewrite $448$ as $8^{2} \cdot 7$ .

Factor $64$ out of $448$ .

$x = \frac{- 8 \pm \sqrt{64 (7)}}{2 \cdot 1}$

Rewrite $64$ as $8^{2}$ .

$x = \frac{- 8 \pm \sqrt{8^{2} \cdot 7}}{2 \cdot 1}$

Pull terms out from under the radical.

$x = \frac{- 8 \pm 8 \sqrt{7}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{- 8 \pm 8 \sqrt{7}}{2}$

Simplify $\frac{- 8 \pm 8 \sqrt{7}}{2}$ .

$x = - 4 \pm 4 \sqrt{7}$

The final answer is the combination of both solutions.

$x = - 4 + 4 \sqrt{7}, - 4 - 4 \sqrt{7}$

$x = - 4 \pm 4 \sqrt{7}$

Step 3

The result can be shown in multiple forms.

Exact Form:

$x = - 4 \pm 4 \sqrt{7}$

Decimal Form:

$x = 6.58300524 \dots, - 14.58300524 \dots$

Step 4