Find the Roots (Zeros) (x^2)/4=2x-10

$\frac{x^{2}}{4} = 2 x - 10$

Multiply both sides of the equation by $4$ .

$x^{2} = 2 x \cdot (4) - 10 \cdot 4$

Simplify each term.

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Multiply $4$ by $2$ .

$x^{2} = 8 x - 10 \cdot 4$

Multiply $- 10$ by $4$ .

$x^{2} = 8 x - 40$

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

$x^{2} - 8 x = - 40$

Move $40$ to the left side of the equation by adding it to both sides.

$x^{2} - 8 x + 40 = 0$

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 1$ , $b = - 8$ , and $c = 40$ into the quadratic formula and solve for $x$ .

$\frac{8 \pm \sqrt{{(- 8)}^{2} - 4 \cdot (1 \cdot 40)}}{2 \cdot 1}$

Simplify.

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Simplify the numerator.

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

$x = \frac{8 \pm \sqrt{64 - 4 \cdot (1 \cdot 40)}}{2 \cdot 1}$

Multiply $40$ by $1$ .

$x = \frac{8 \pm \sqrt{64 - 4 \cdot 40}}{2 \cdot 1}$

Multiply $- 4$ by $40$ .

$x = \frac{8 \pm \sqrt{64 - 160}}{2 \cdot 1}$

Subtract $160$ from $64$ .

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

Rewrite $- 96$ as $- 1 (96)$ .

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

Rewrite $\sqrt{- 1 (96)}$ as $\sqrt{- 1} \cdot \sqrt{96}$ .

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

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

$x = \frac{8 \pm i \cdot \sqrt{96}}{2 \cdot 1}$

Rewrite $96$ as $4^{2} \cdot 6$ .

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Factor $16$ out of $96$ .

$x = \frac{8 \pm i \cdot \sqrt{16 (6)}}{2 \cdot 1}$

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

$x = \frac{8 \pm i \cdot \sqrt{4^{2} \cdot 6}}{2 \cdot 1}$

Pull terms out from under the radical.

$x = \frac{8 \pm i \cdot (4 \sqrt{6})}{2 \cdot 1}$

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

$x = \frac{8 \pm 4 i \sqrt{6}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{8 \pm 4 i \sqrt{6}}{2}$

Simplify $\frac{8 \pm 4 i \sqrt{6}}{2}$ .

$x = 4 \pm 2 i \sqrt{6}$

Find the Roots (Zeros) (x^2)/4=2x-10

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