Solve Using the Quadratic Formula 5x^2+4x=-4

$5 x^{2} + 4 x = - 4$

Step 1

Add $4$ to both sides of the equation.

$5 x^{2} + 4 x + 4 = 0$

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 $a = 5$ , $b = 4$ , and $c = 4$ into the quadratic formula and solve for $x$ .

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

Step 4

Simplify.

Simplify the numerator.

Raise $4$ to the power of $2$ .

$x = \frac{- 4 \pm \sqrt{16 - 4 \cdot 5 \cdot 4}}{2 \cdot 5}$

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

Multiply $- 4$ by $5$ .

$x = \frac{- 4 \pm \sqrt{16 - 20 \cdot 4}}{2 \cdot 5}$

Multiply $- 20$ by $4$ .

$x = \frac{- 4 \pm \sqrt{16 - 80}}{2 \cdot 5}$

Subtract $80$ from $16$ .

$x = \frac{- 4 \pm \sqrt{- 64}}{2 \cdot 5}$

Rewrite $- 64$ as $- 1 (64)$ .

$x = \frac{- 4 \pm \sqrt{- 1 \cdot 64}}{2 \cdot 5}$

Rewrite $\sqrt{- 1 (64)}$ as $\sqrt{- 1} \cdot \sqrt{64}$ .

$x = \frac{- 4 \pm \sqrt{- 1} \cdot \sqrt{64}}{2 \cdot 5}$

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

$x = \frac{- 4 \pm i \cdot \sqrt{64}}{2 \cdot 5}$

Rewrite $64$ as $8^{2}$ .

$x = \frac{- 4 \pm i \cdot \sqrt{8^{2}}}{2 \cdot 5}$

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

$x = \frac{- 4 \pm i \cdot 8}{2 \cdot 5}$

Move $8$ to the left of $i$ .

$x = \frac{- 4 \pm 8 i}{2 \cdot 5}$

Multiply $2$ by $5$ .

$x = \frac{- 4 \pm 8 i}{10}$

Simplify $\frac{- 4 \pm 8 i}{10}$ .

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