Solve Using the Quadratic Formula 2x^2-5x-7=0

$2 x^{2} - 5 x - 7 = 0$

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 $a = 2$ , $b = - 5$ , and $c = - 7$ into the quadratic formula and solve for $x$ .

$\frac{5 \pm \sqrt{{(- 5)}^{2} - 4 \cdot (2 \cdot - 7)}}{2 \cdot 2}$

Step 3

Simplify.

Simplify the numerator.

Raise $- 5$ to the power of $2$ .

$x = \frac{5 \pm \sqrt{25 - 4 \cdot 2 \cdot - 7}}{2 \cdot 2}$

Multiply $- 4 \cdot 2 \cdot - 7$ .

Multiply $- 4$ by $2$ .

$x = \frac{5 \pm \sqrt{25 - 8 \cdot - 7}}{2 \cdot 2}$

Multiply $- 8$ by $- 7$ .

$x = \frac{5 \pm \sqrt{25 + 56}}{2 \cdot 2}$

Add $25$ and $56$ .

$x = \frac{5 \pm \sqrt{81}}{2 \cdot 2}$

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

$x = \frac{5 \pm \sqrt{9^{2}}}{2 \cdot 2}$

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

$x = \frac{5 \pm 9}{2 \cdot 2}$

Multiply $2$ by $2$ .

$x = \frac{5 \pm 9}{4}$

Step 4

The final answer is the combination of both solutions.

$x = \frac{7}{2}, - 1$