Solve Using the Quadratic Formula 3x^2+6x-9=0

$3 x^{2} + 6 x - 9 = 0$

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 $a = 3$ , $b = 6$ , and $c = - 9$ into the quadratic formula and solve for $x$ .

$\frac{- 6 \pm \sqrt{6^{2} - 4 \cdot (3 \cdot - 9)}}{2 \cdot 3}$

Step 3

Simplify.

Simplify the numerator.

Raise $6$ to the power of $2$ .

$x = \frac{- 6 \pm \sqrt{36 - 4 \cdot 3 \cdot - 9}}{2 \cdot 3}$

Multiply $- 4 \cdot 3 \cdot - 9$ .

Multiply $- 4$ by $3$ .

$x = \frac{- 6 \pm \sqrt{36 - 12 \cdot - 9}}{2 \cdot 3}$

Multiply $- 12$ by $- 9$ .

$x = \frac{- 6 \pm \sqrt{36 + 108}}{2 \cdot 3}$

Add $36$ and $108$ .

$x = \frac{- 6 \pm \sqrt{144}}{2 \cdot 3}$

Rewrite $144$ as $12^{2}$ .

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

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

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

Multiply $2$ by $3$ .

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

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

$x = - 1 \pm 2$

Step 4

The final answer is the combination of both solutions.

$x = 1, - 3$