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

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

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 $a = 8$ , $b = - 2$ , and $c = - 5$ into the quadratic formula and solve for $x$ .

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

Step 3

Simplify.

Simplify the numerator.

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

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

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

Multiply $- 4$ by $8$ .

$x = \frac{2 \pm \sqrt{4 - 32 \cdot - 5}}{2 \cdot 8}$

Multiply $- 32$ by $- 5$ .

$x = \frac{2 \pm \sqrt{4 + 160}}{2 \cdot 8}$

Add $4$ and $160$ .

$x = \frac{2 \pm \sqrt{164}}{2 \cdot 8}$

Rewrite $164$ as $2^{2} \cdot 41$ .

Factor $4$ out of $164$ .

$x = \frac{2 \pm \sqrt{4 (41)}}{2 \cdot 8}$

Rewrite $4$ as $2^{2}$ .

$x = \frac{2 \pm \sqrt{2^{2} \cdot 41}}{2 \cdot 8}$

Pull terms out from under the radical.

$x = \frac{2 \pm 2 \sqrt{41}}{2 \cdot 8}$

Multiply $2$ by $8$ .

$x = \frac{2 \pm 2 \sqrt{41}}{16}$

Simplify $\frac{2 \pm 2 \sqrt{41}}{16}$ .

$x = \frac{1 \pm \sqrt{41}}{8}$

Step 4

The result can be shown in multiple forms.

Exact Form:

$x = \frac{1 \pm \sqrt{41}}{8}$

Decimal Form:

$x = 0.92539052 \dots, - 0.67539052 \dots$