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

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

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

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

Step 3

Simplify.

Simplify the numerator.

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

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

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

Multiply $- 4$ by $2$ .

$x = \frac{3 \pm \sqrt{9 - 8 \cdot - 9}}{2 \cdot 2}$

Multiply $- 8$ by $- 9$ .

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

Add $9$ and $72$ .

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

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

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

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

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

Multiply $2$ by $2$ .

$x = \frac{3 \pm 9}{4}$

Step 4

The final answer is the combination of both solutions.

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