Solve Using the Quadratic Formula 8x^2-6x-5=0

$8 x^{2} - 6 x - 5 = 0$

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 $a = 8$ , $b = - 6$ , and $c = - 5$ into the quadratic formula and solve for $x$ .

$\frac{6 \pm \sqrt{{(- 6)}^{2} - 4 \cdot (8 \cdot - 5)}}{2 \cdot 8}$

Step 3

Simplify.

Simplify the numerator.

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

$x = \frac{6 \pm \sqrt{36 - 4 \cdot 8 \cdot - 5}}{2 \cdot 8}$

Multiply $- 4 \cdot 8 \cdot - 5$ .

Multiply $- 4$ by $8$ .

$x = \frac{6 \pm \sqrt{36 - 32 \cdot - 5}}{2 \cdot 8}$

Multiply $- 32$ by $- 5$ .

$x = \frac{6 \pm \sqrt{36 + 160}}{2 \cdot 8}$

Add $36$ and $160$ .

$x = \frac{6 \pm \sqrt{196}}{2 \cdot 8}$

Rewrite $196$ as $14^{2}$ .

$x = \frac{6 \pm \sqrt{14^{2}}}{2 \cdot 8}$

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

$x = \frac{6 \pm 14}{2 \cdot 8}$

Multiply $2$ by $8$ .

$x = \frac{6 \pm 14}{16}$

Simplify $\frac{6 \pm 14}{16}$ .

$x = \frac{3 \pm 7}{8}$

Step 4

The final answer is the combination of both solutions.

$x = \frac{5}{4}, - \frac{1}{2}$