Solve Using the Quadratic Formula 5x^2=2x-5

$5 x^{2} = 2 x - 5$

Step 1

Move all the expressions to the left side of the equation.

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

$5 x^{2} - 2 x = - 5$

Add $5$ to both sides of the equation.

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

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 $c = 5$ into the quadratic formula and solve for $x$ .

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

Step 4

Simplify.

Simplify the numerator.

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

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

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

Multiply $- 4$ by $5$ .

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

Multiply $- 20$ by $5$ .

$x = \frac{2 \pm \sqrt{4 - 100}}{2 \cdot 5}$

Subtract $100$ from $4$ .

$x = \frac{2 \pm \sqrt{- 96}}{2 \cdot 5}$

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

$x = \frac{2 \pm \sqrt{- 1 \cdot 96}}{2 \cdot 5}$

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

$x = \frac{2 \pm \sqrt{- 1} \cdot \sqrt{96}}{2 \cdot 5}$

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

$x = \frac{2 \pm i \cdot \sqrt{96}}{2 \cdot 5}$

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

Factor $16$ out of $96$ .

$x = \frac{2 \pm i \cdot \sqrt{16 (6)}}{2 \cdot 5}$

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

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

Pull terms out from under the radical.

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

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

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

Multiply $2$ by $5$ .

$x = \frac{2 \pm 4 i \sqrt{6}}{10}$

Simplify $\frac{2 \pm 4 i \sqrt{6}}{10}$ .

$x = \frac{1 \pm 2 i \sqrt{6}}{5}$