Find All Complex Solutions 5x^2+10=2x

$5 x^{2} + 10 = 2 x$

Step 1

Subtract $2 x$ from both sides of the equation.

$5 x^{2} + 10 - 2 x = 0$

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 $a = 5$ , $b = - 2$ , and $c = 10$ into the quadratic formula and solve for $x$ .

$\frac{2 \pm \sqrt{{(- 2)}^{2} - 4 \cdot (5 \cdot 10)}}{2 \cdot 5}$

Step 4

Simplify.

Simplify the numerator.

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

$x = \frac{2 \pm \sqrt{4 - 4 \cdot 5 \cdot 10}}{2 \cdot 5}$

Multiply $- 4 \cdot 5 \cdot 10$ .

Multiply $- 4$ by $5$ .

$x = \frac{2 \pm \sqrt{4 - 20 \cdot 10}}{2 \cdot 5}$

Multiply $- 20$ by $10$ .

$x = \frac{2 \pm \sqrt{4 - 200}}{2 \cdot 5}$

Subtract $200$ from $4$ .

$x = \frac{2 \pm \sqrt{- 196}}{2 \cdot 5}$

Rewrite $- 196$ as $- 1 (196)$ .

$x = \frac{2 \pm \sqrt{- 1 \cdot 196}}{2 \cdot 5}$

Rewrite $\sqrt{- 1 (196)}$ as $\sqrt{- 1} \cdot \sqrt{196}$ .

$x = \frac{2 \pm \sqrt{- 1} \cdot \sqrt{196}}{2 \cdot 5}$

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

$x = \frac{2 \pm i \cdot \sqrt{196}}{2 \cdot 5}$

Rewrite $196$ as $14^{2}$ .

$x = \frac{2 \pm i \cdot \sqrt{14^{2}}}{2 \cdot 5}$

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

$x = \frac{2 \pm i \cdot 14}{2 \cdot 5}$

Move $14$ to the left of $i$ .

$x = \frac{2 \pm 14 i}{2 \cdot 5}$

Multiply $2$ by $5$ .

$x = \frac{2 \pm 14 i}{10}$

Simplify $\frac{2 \pm 14 i}{10}$ .

$x = \frac{1 \pm 7 i}{5}$