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

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

Step 1

Add $2$ to both sides of the equation.

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

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 $c = 2$ into the quadratic formula and solve for $x$ .

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

Step 4

Simplify.

Simplify the numerator.

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

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

Multiply $- 4 \cdot 5 \cdot 2$ .

Multiply $- 4$ by $5$ .

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

Multiply $- 20$ by $2$ .

$x = \frac{- 2 \pm \sqrt{4 - 40}}{2 \cdot 5}$

Subtract $40$ from $4$ .

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

Rewrite $- 36$ as $- 1 (36)$ .

$x = \frac{- 2 \pm \sqrt{- 1 \cdot 36}}{2 \cdot 5}$

Rewrite $\sqrt{- 1 (36)}$ as $\sqrt{- 1} \cdot \sqrt{36}$ .

$x = \frac{- 2 \pm \sqrt{- 1} \cdot \sqrt{36}}{2 \cdot 5}$

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

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

Rewrite $36$ as $6^{2}$ .

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

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

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

Move $6$ to the left of $i$ .

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

Multiply $2$ by $5$ .

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

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

$x = \frac{- 1 \pm 3 i}{5}$