Solve Using the Quadratic Formula 8x^2-2x-1=0

$8 x^{2} - 2 x - 1 = 0$

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 $c = - 1$ into the quadratic formula and solve for $x$ .

$\frac{2 \pm \sqrt{{(- 2)}^{2} - 4 \cdot (8 \cdot - 1)}}{2 \cdot 8}$

Step 3

Simplify.

Simplify the numerator.

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

$x = \frac{2 \pm \sqrt{4 - 4 \cdot 8 \cdot - 1}}{2 \cdot 8}$

Multiply $- 4 \cdot 8 \cdot - 1$ .

Multiply $- 4$ by $8$ .

$x = \frac{2 \pm \sqrt{4 - 32 \cdot - 1}}{2 \cdot 8}$

Multiply $- 32$ by $- 1$ .

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

Add $4$ and $32$ .

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

Rewrite $36$ as $6^{2}$ .

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

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

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

Multiply $2$ by $8$ .

$x = \frac{2 \pm 6}{16}$

Simplify $\frac{2 \pm 6}{16}$ .

$x = \frac{1 \pm 3}{8}$

Step 4

The final answer is the combination of both solutions.

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