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

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

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 $a = 9$ , $b = - 3$ , and $c = - 2$ into the quadratic formula and solve for $x$ .

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

Step 3

Simplify.

Simplify the numerator.

Raise $- 3$ to the power of $2$ .

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

Multiply $- 4 \cdot 9 \cdot - 2$ .

Multiply $- 4$ by $9$ .

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

Multiply $- 36$ by $- 2$ .

$x = \frac{3 \pm \sqrt{9 + 72}}{2 \cdot 9}$

Add $9$ and $72$ .

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

Rewrite $81$ as $9^{2}$ .

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

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

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

Multiply $2$ by $9$ .

$x = \frac{3 \pm 9}{18}$

Simplify $\frac{3 \pm 9}{18}$ .

$x = \frac{1 \pm 3}{6}$

Step 4

The final answer is the combination of both solutions.

$x = \frac{2}{3}, - \frac{1}{3}$