Solve Using the Quadratic Formula 2x^2-3x-20=0

$2 x^{2} - 3 x - 20 = 0$

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 $c = - 20$ into the quadratic formula and solve for $x$ .

$\frac{3 \pm \sqrt{{(- 3)}^{2} - 4 \cdot (2 \cdot - 20)}}{2 \cdot 2}$

Step 3

Simplify.

Simplify the numerator.

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

$x = \frac{3 \pm \sqrt{9 - 4 \cdot 2 \cdot - 20}}{2 \cdot 2}$

Multiply $- 4 \cdot 2 \cdot - 20$ .

Multiply $- 4$ by $2$ .

$x = \frac{3 \pm \sqrt{9 - 8 \cdot - 20}}{2 \cdot 2}$

Multiply $- 8$ by $- 20$ .

$x = \frac{3 \pm \sqrt{9 + 160}}{2 \cdot 2}$

Add $9$ and $160$ .

$x = \frac{3 \pm \sqrt{169}}{2 \cdot 2}$

Rewrite $169$ as $13^{2}$ .

$x = \frac{3 \pm \sqrt{13^{2}}}{2 \cdot 2}$

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

$x = \frac{3 \pm 13}{2 \cdot 2}$

Multiply $2$ by $2$ .

$x = \frac{3 \pm 13}{4}$

Step 4

The final answer is the combination of both solutions.

$x = 4, - \frac{5}{2}$