Solve Using the Quadratic Formula -3x^2-4x-4=0

$- 3 x^{2} - 4 x - 4 = 0$

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 $a = - 3$ , $b = - 4$ , and $c = - 4$ into the quadratic formula and solve for $x$ .

$\frac{4 \pm \sqrt{{(- 4)}^{2} - 4 \cdot (- 3 \cdot - 4)}}{2 \cdot - 3}$

Step 3

Simplify.

Simplify the numerator.

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

$x = \frac{4 \pm \sqrt{16 - 4 \cdot - 3 \cdot - 4}}{2 \cdot - 3}$

Multiply $- 4 \cdot - 3 \cdot - 4$ .

Multiply $- 4$ by $- 3$ .

$x = \frac{4 \pm \sqrt{16 + 12 \cdot - 4}}{2 \cdot - 3}$

Multiply $12$ by $- 4$ .

$x = \frac{4 \pm \sqrt{16 - 48}}{2 \cdot - 3}$

Subtract $48$ from $16$ .

$x = \frac{4 \pm \sqrt{- 32}}{2 \cdot - 3}$

Rewrite $- 32$ as $- 1 (32)$ .

$x = \frac{4 \pm \sqrt{- 1 \cdot 32}}{2 \cdot - 3}$

Rewrite $\sqrt{- 1 (32)}$ as $\sqrt{- 1} \cdot \sqrt{32}$ .

$x = \frac{4 \pm \sqrt{- 1} \cdot \sqrt{32}}{2 \cdot - 3}$

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

$x = \frac{4 \pm i \cdot \sqrt{32}}{2 \cdot - 3}$

Rewrite $32$ as $4^{2} \cdot 2$ .

Factor $16$ out of $32$ .

$x = \frac{4 \pm i \cdot \sqrt{16 (2)}}{2 \cdot - 3}$

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

$x = \frac{4 \pm i \cdot \sqrt{4^{2} \cdot 2}}{2 \cdot - 3}$

Pull terms out from under the radical.

$x = \frac{4 \pm i \cdot (4 \sqrt{2})}{2 \cdot - 3}$

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

$x = \frac{4 \pm 4 i \sqrt{2}}{2 \cdot - 3}$

Multiply $2$ by $- 3$ .

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

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

$x = \frac{2 \pm 2 i \sqrt{2}}{- 3}$

Move the negative in front of the fraction.

$x = - \frac{2 \pm 2 i \sqrt{2}}{3}$