Solve Using the Quadratic Formula 5x^2+12x+8=0

$5 x^{2} + 12 x + 8 = 0$

Step 1

Use the quadratic formula to find the solutions.

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

Step 2

Substitute the values $a = 5$ , $b = 12$ , and $c = 8$ into the quadratic formula and solve for $x$ .

$\frac{- 12 \pm \sqrt{12^{2} - 4 \cdot (5 \cdot 8)}}{2 \cdot 5}$

Step 3

Simplify.

Simplify the numerator.

Raise $12$ to the power of $2$ .

$x = \frac{- 12 \pm \sqrt{144 - 4 \cdot 5 \cdot 8}}{2 \cdot 5}$

Multiply $- 4 \cdot 5 \cdot 8$ .

Multiply $- 4$ by $5$ .

$x = \frac{- 12 \pm \sqrt{144 - 20 \cdot 8}}{2 \cdot 5}$

Multiply $- 20$ by $8$ .

$x = \frac{- 12 \pm \sqrt{144 - 160}}{2 \cdot 5}$

Subtract $160$ from $144$ .

$x = \frac{- 12 \pm \sqrt{- 16}}{2 \cdot 5}$

Rewrite $- 16$ as $- 1 (16)$ .

$x = \frac{- 12 \pm \sqrt{- 1 \cdot 16}}{2 \cdot 5}$

Rewrite $\sqrt{- 1 (16)}$ as $\sqrt{- 1} \cdot \sqrt{16}$ .

$x = \frac{- 12 \pm \sqrt{- 1} \cdot \sqrt{16}}{2 \cdot 5}$

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

$x = \frac{- 12 \pm i \cdot \sqrt{16}}{2 \cdot 5}$

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

$x = \frac{- 12 \pm i \cdot \sqrt{4^{2}}}{2 \cdot 5}$

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

$x = \frac{- 12 \pm i \cdot 4}{2 \cdot 5}$

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

$x = \frac{- 12 \pm 4 i}{2 \cdot 5}$

Multiply $2$ by $5$ .

$x = \frac{- 12 \pm 4 i}{10}$

Simplify $\frac{- 12 \pm 4 i}{10}$ .

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