Solve Using the Quadratic Formula 3x^2+9x-12=0

$3 x^{2} + 9 x - 12 = 0$

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 $a = 3$ , $b = 9$ , and $c = - 12$ into the quadratic formula and solve for $x$ .

$\frac{- 9 \pm \sqrt{9^{2} - 4 \cdot (3 \cdot - 12)}}{2 \cdot 3}$

Step 3

Simplify.

Simplify the numerator.

Raise $9$ to the power of $2$ .

$x = \frac{- 9 \pm \sqrt{81 - 4 \cdot 3 \cdot - 12}}{2 \cdot 3}$

Multiply $- 4 \cdot 3 \cdot - 12$ .

Multiply $- 4$ by $3$ .

$x = \frac{- 9 \pm \sqrt{81 - 12 \cdot - 12}}{2 \cdot 3}$

Multiply $- 12$ by $- 12$ .

$x = \frac{- 9 \pm \sqrt{81 + 144}}{2 \cdot 3}$

Add $81$ and $144$ .

$x = \frac{- 9 \pm \sqrt{225}}{2 \cdot 3}$

Rewrite $225$ as $15^{2}$ .

$x = \frac{- 9 \pm \sqrt{15^{2}}}{2 \cdot 3}$

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

$x = \frac{- 9 \pm 15}{2 \cdot 3}$

Multiply $2$ by $3$ .

$x = \frac{- 9 \pm 15}{6}$

Simplify $\frac{- 9 \pm 15}{6}$ .

$x = \frac{- 3 \pm 5}{2}$

Step 4

The final answer is the combination of both solutions.

$x = 1, - 4$