Find the Roots (Zeros) x^2+12=4x

$x^{2} + 12 = 4 x$

Step 1

Subtract $4 x$ from both sides of the equation.

$x^{2} + 12 - 4 x = 0$

Step 2

Use the quadratic formula to find the solutions.

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

Step 3

Substitute the values $a = 1$ , $b = - 4$ , and $c = 12$ into the quadratic formula and solve for $x$ .

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

Step 4

Simplify.

Simplify the numerator.

Raise $- 4$ to the power of $2$ .

$x = \frac{4 \pm \sqrt{16 - 4 \cdot 1 \cdot 12}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot 12$ .

Multiply $- 4$ by $1$ .

$x = \frac{4 \pm \sqrt{16 - 4 \cdot 12}}{2 \cdot 1}$

Multiply $- 4$ by $12$ .

$x = \frac{4 \pm \sqrt{16 - 48}}{2 \cdot 1}$

Subtract $48$ from $16$ .

$x = \frac{4 \pm \sqrt{- 32}}{2 \cdot 1}$

Rewrite $- 32$ as $- 1 (32)$ .

$x = \frac{4 \pm \sqrt{- 1 \cdot 32}}{2 \cdot 1}$

Rewrite $\sqrt{- 1 (32)}$ as $\sqrt{- 1} \cdot \sqrt{32}$ .

$x = \frac{4 \pm \sqrt{- 1} \cdot \sqrt{32}}{2 \cdot 1}$

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \frac{4 \pm i \cdot \sqrt{32}}{2 \cdot 1}$

Rewrite $32$ as $4^{2} \cdot 2$ .

Factor $16$ out of $32$ .

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

Rewrite $16$ as $4^{2}$ .

$x = \frac{4 \pm i \cdot \sqrt{4^{2} \cdot 2}}{2 \cdot 1}$

Pull terms out from under the radical.

$x = \frac{4 \pm i \cdot (4 \sqrt{2})}{2 \cdot 1}$

Move $4$ to the left of $i$ .

$x = \frac{4 \pm 4 i \sqrt{2}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{4 \pm 4 i \sqrt{2}}{2}$

Simplify $\frac{4 \pm 4 i \sqrt{2}}{2}$ .

$x = 2 \pm 2 i \sqrt{2}$

Step 5