Solve Using the Quadratic Formula 9x^2-6x-8=0

$9 x^{2} - 6 x - 8 = 0$

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 $a = 9$ , $b = - 6$ , and $c = - 8$ into the quadratic formula and solve for $x$ .

$\frac{6 \pm \sqrt{{(- 6)}^{2} - 4 \cdot (9 \cdot - 8)}}{2 \cdot 9}$

Step 3

Simplify.

Simplify the numerator.

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

$x = \frac{6 \pm \sqrt{36 - 4 \cdot 9 \cdot - 8}}{2 \cdot 9}$

Multiply $- 4 \cdot 9 \cdot - 8$ .

Multiply $- 4$ by $9$ .

$x = \frac{6 \pm \sqrt{36 - 36 \cdot - 8}}{2 \cdot 9}$

Multiply $- 36$ by $- 8$ .

$x = \frac{6 \pm \sqrt{36 + 288}}{2 \cdot 9}$

Add $36$ and $288$ .

$x = \frac{6 \pm \sqrt{324}}{2 \cdot 9}$

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

$x = \frac{6 \pm \sqrt{18^{2}}}{2 \cdot 9}$

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

$x = \frac{6 \pm 18}{2 \cdot 9}$

Multiply $2$ by $9$ .

$x = \frac{6 \pm 18}{18}$

Simplify $\frac{6 \pm 18}{18}$ .

$x = \frac{1 \pm 3}{3}$

Step 4

The final answer is the combination of both solutions.

$x = \frac{4}{3}, - \frac{2}{3}$