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

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

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

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

Step 3

Simplify.

Simplify the numerator.

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

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

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

Multiply $- 4$ by $5$ .

$x = \frac{12 \pm \sqrt{144 - 20 \cdot - 9}}{2 \cdot 5}$

Multiply $- 20$ by $- 9$ .

$x = \frac{12 \pm \sqrt{144 + 180}}{2 \cdot 5}$

Add $144$ and $180$ .

$x = \frac{12 \pm \sqrt{324}}{2 \cdot 5}$

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

$x = \frac{12 \pm \sqrt{18^{2}}}{2 \cdot 5}$

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

$x = \frac{12 \pm 18}{2 \cdot 5}$

Multiply $2$ by $5$ .

$x = \frac{12 \pm 18}{10}$

Simplify $\frac{12 \pm 18}{10}$ .

$x = \frac{6 \pm 9}{5}$

Step 4

The final answer is the combination of both solutions.

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