Solve Using the Quadratic Formula 3x^2=-12x-15

$3 x^{2} = - 12 x - 15$

Step 1

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

Add $12 x$ to both sides of the equation.

$3 x^{2} + 12 x = - 15$

Add $15$ to both sides of the equation.

$3 x^{2} + 12 x + 15 = 0$

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 $a = 3$ , $b = 12$ , and $c = 15$ into the quadratic formula and solve for $x$ .

$\frac{- 12 \pm \sqrt{12^{2} - 4 \cdot (3 \cdot 15)}}{2 \cdot 3}$

Step 4

Simplify.

Simplify the numerator.

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

$x = \frac{- 12 \pm \sqrt{144 - 4 \cdot 3 \cdot 15}}{2 \cdot 3}$

Multiply $- 4 \cdot 3 \cdot 15$ .

Multiply $- 4$ by $3$ .

$x = \frac{- 12 \pm \sqrt{144 - 12 \cdot 15}}{2 \cdot 3}$

Multiply $- 12$ by $15$ .

$x = \frac{- 12 \pm \sqrt{144 - 180}}{2 \cdot 3}$

Subtract $180$ from $144$ .

$x = \frac{- 12 \pm \sqrt{- 36}}{2 \cdot 3}$

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

$x = \frac{- 12 \pm \sqrt{- 1 \cdot 36}}{2 \cdot 3}$

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

$x = \frac{- 12 \pm \sqrt{- 1} \cdot \sqrt{36}}{2 \cdot 3}$

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

$x = \frac{- 12 \pm i \cdot \sqrt{36}}{2 \cdot 3}$

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

$x = \frac{- 12 \pm i \cdot \sqrt{6^{2}}}{2 \cdot 3}$

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

$x = \frac{- 12 \pm i \cdot 6}{2 \cdot 3}$

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

$x = \frac{- 12 \pm 6 i}{2 \cdot 3}$

Multiply $2$ by $3$ .

$x = \frac{- 12 \pm 6 i}{6}$

Simplify $\frac{- 12 \pm 6 i}{6}$ .

$x = - 2 \pm i$