Find All Complex Solutions x^2=36

$x^{2} = 36$

Subtract $36$ from both sides of the equation.

$x^{2} - 36 = 0$

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 1$ , $b = 0$ , and $c = - 36$ into the quadratic formula and solve for $x$ .

$\frac{0 \pm \sqrt{0^{2} - 4 \cdot (1 \cdot - 36)}}{2 \cdot 1}$

Simplify.

Simplify the numerator.

Raising $0$ to any positive power yields $0$ .

$x = \frac{0 \pm \sqrt{0 - 4 \cdot 1 \cdot - 36}}{2 \cdot 1}$

Multiply $- 4$ by $1$ .

$x = \frac{0 \pm \sqrt{0 - 4 \cdot - 36}}{2 \cdot 1}$

Multiply $- 4$ by $- 36$ .

$x = \frac{0 \pm \sqrt{0 + 144}}{2 \cdot 1}$

Add $0$ and $144$ .

$x = \frac{0 \pm \sqrt{144}}{2 \cdot 1}$

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

$x = \frac{0 \pm \sqrt{12^{2}}}{2 \cdot 1}$

Pull terms out from under the radical, assuming positive real numbers.

$x = \frac{0 \pm 12}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{0 \pm 12}{2}$

Simplify $\frac{0 \pm 12}{2}$ .

$x = \pm 6$

The final answer is the combination of both solutions.

$x = 6, - 6$