Find the Roots (Zeros) f(x)=x^2+12x-36

$f (x) = x^{2} + 12 x - 36$

Step 1

Set $x^{2} + 12 x - 36$ equal to $0$ .

$x^{2} + 12 x - 36 = 0$

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 $a = 1$ , $b = 12$ , and $c = - 36$ into the quadratic formula and solve for $x$ .

$\frac{- 12 \pm \sqrt{12^{2} - 4 \cdot (1 \cdot - 36)}}{2 \cdot 1}$

Simplify.

Simplify the numerator.

Raise $12$ to the power of $2$ .

$x = \frac{- 12 \pm \sqrt{144 - 4 \cdot 1 \cdot - 36}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot - 36$ .

Multiply $- 4$ by $1$ .

$x = \frac{- 12 \pm \sqrt{144 - 4 \cdot - 36}}{2 \cdot 1}$

Multiply $- 4$ by $- 36$ .

$x = \frac{- 12 \pm \sqrt{144 + 144}}{2 \cdot 1}$

Add $144$ and $144$ .

$x = \frac{- 12 \pm \sqrt{288}}{2 \cdot 1}$

Rewrite $288$ as $12^{2} \cdot 2$ .

Factor $144$ out of $288$ .

$x = \frac{- 12 \pm \sqrt{144 (2)}}{2 \cdot 1}$

Rewrite $144$ as $12^{2}$ .

$x = \frac{- 12 \pm \sqrt{12^{2} \cdot 2}}{2 \cdot 1}$

Pull terms out from under the radical.

$x = \frac{- 12 \pm 12 \sqrt{2}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{- 12 \pm 12 \sqrt{2}}{2}$

Simplify $\frac{- 12 \pm 12 \sqrt{2}}{2}$ .

$x = - 6 \pm 6 \sqrt{2}$

The final answer is the combination of both solutions.

$x = - 6 + 6 \sqrt{2}, - 6 - 6 \sqrt{2}$

$x = - 6 \pm 6 \sqrt{2}$

Step 3

The result can be shown in multiple forms.

Exact Form:

$x = - 6 \pm 6 \sqrt{2}$

Decimal Form:

$x = 2.48528137 \dots, - 14.48528137 \dots$

Step 4