Find the Roots (Zeros) x^2=8x+4

$x^{2} = 8 x + 4$

Step 1

Subtract $8 x$ from both sides of the equation.

$x^{2} - 8 x = 4$

Step 2

Subtract $4$ from both sides of the equation.

$x^{2} - 8 x - 4 = 0$

Step 3

Use the quadratic formula to find the solutions.

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

Step 4

Substitute the values $a = 1$ , $b = - 8$ , and $c = - 4$ into the quadratic formula and solve for $x$ .

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

Step 5

Simplify.

Simplify the numerator.

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

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

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

Multiply $- 4$ by $1$ .

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

Multiply $- 4$ by $- 4$ .

$x = \frac{8 \pm \sqrt{64 + 16}}{2 \cdot 1}$

Add $64$ and $16$ .

$x = \frac{8 \pm \sqrt{80}}{2 \cdot 1}$

Rewrite $80$ as $4^{2} \cdot 5$ .

Factor $16$ out of $80$ .

$x = \frac{8 \pm \sqrt{16 (5)}}{2 \cdot 1}$

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

$x = \frac{8 \pm \sqrt{4^{2} \cdot 5}}{2 \cdot 1}$

Pull terms out from under the radical.

$x = \frac{8 \pm 4 \sqrt{5}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{8 \pm 4 \sqrt{5}}{2}$

Simplify $\frac{8 \pm 4 \sqrt{5}}{2}$ .

$x = 4 \pm 2 \sqrt{5}$

Step 6

The result can be shown in multiple forms.

Exact Form:

$x = 4 \pm 2 \sqrt{5}$

Decimal Form:

$x = 8.47213595 \dots, - 0.47213595 \dots$

Step 7