Solve Using the Quadratic Formula 9x^2+12x-5=0

$9 x^{2} + 12 x - 5 = 0$

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 $a = 9$ , $b = 12$ , and $c = - 5$ into the quadratic formula and solve for $x$ .

$\frac{- 12 \pm \sqrt{12^{2} - 4 \cdot (9 \cdot - 5)}}{2 \cdot 9}$

Step 3

Simplify.

Simplify the numerator.

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

$x = \frac{- 12 \pm \sqrt{144 - 4 \cdot 9 \cdot - 5}}{2 \cdot 9}$

Multiply $- 4 \cdot 9 \cdot - 5$ .

Multiply $- 4$ by $9$ .

$x = \frac{- 12 \pm \sqrt{144 - 36 \cdot - 5}}{2 \cdot 9}$

Multiply $- 36$ by $- 5$ .

$x = \frac{- 12 \pm \sqrt{144 + 180}}{2 \cdot 9}$

Add $144$ and $180$ .

$x = \frac{- 12 \pm \sqrt{324}}{2 \cdot 9}$

Rewrite $324$ as $18^{2}$ .

$x = \frac{- 12 \pm \sqrt{18^{2}}}{2 \cdot 9}$

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

$x = \frac{- 12 \pm 18}{2 \cdot 9}$

Multiply $2$ by $9$ .

$x = \frac{- 12 \pm 18}{18}$

Simplify $\frac{- 12 \pm 18}{18}$ .

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

Step 4

The final answer is the combination of both solutions.

$x = \frac{1}{3}, - \frac{5}{3}$