Find All Complex Solutions 5x^2-6x+2=0

$5 x^{2} - 6 x + 2 = 0$

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 5$ , $b = - 6$ , and $c = 2$ into the quadratic formula and solve for $x$ .

$\frac{6 \pm \sqrt{{(- 6)}^{2} - 4 \cdot (5 \cdot 2)}}{2 \cdot 5}$

Simplify.

Simplify the numerator.

Raise $- 6$ to the power of $2$ .

$x = \frac{6 \pm \sqrt{36 - 4 \cdot 5 \cdot 2}}{2 \cdot 5}$

Multiply $- 4$ by $5$ .

$x = \frac{6 \pm \sqrt{36 - 20 \cdot 2}}{2 \cdot 5}$

Multiply $- 20$ by $2$ .

$x = \frac{6 \pm \sqrt{36 - 40}}{2 \cdot 5}$

Subtract $40$ from $36$ .

$x = \frac{6 \pm \sqrt{- 4}}{2 \cdot 5}$

Rewrite $- 4$ as $- 1 (4)$ .

$x = \frac{6 \pm \sqrt{- 1 \cdot 4}}{2 \cdot 5}$

Rewrite $\sqrt{- 1 (4)}$ as $\sqrt{- 1} \cdot \sqrt{4}$ .

$x = \frac{6 \pm \sqrt{- 1} \cdot \sqrt{4}}{2 \cdot 5}$

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

$x = \frac{6 \pm i \cdot \sqrt{4}}{2 \cdot 5}$

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

$x = \frac{6 \pm i \cdot \sqrt{2^{2}}}{2 \cdot 5}$

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

$x = \frac{6 \pm i \cdot 2}{2 \cdot 5}$

Move $2$ to the left of $i$ .

$x = \frac{6 \pm 2 i}{2 \cdot 5}$

Multiply $2$ by $5$ .

$x = \frac{6 \pm 2 i}{10}$

Simplify $\frac{6 \pm 2 i}{10}$ .

$x = \frac{3 \pm i}{5}$