Solve Using the Quadratic Formula 8x^2+4x-4=0

$8 x^{2} + 4 x - 4 = 0$

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 $a = 8$ , $b = 4$ , and $c = - 4$ into the quadratic formula and solve for $x$ .

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

Step 3

Simplify.

Simplify the numerator.

Raise $4$ to the power of $2$ .

$x = \frac{- 4 \pm \sqrt{16 - 4 \cdot 8 \cdot - 4}}{2 \cdot 8}$

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

Multiply $- 4$ by $8$ .

$x = \frac{- 4 \pm \sqrt{16 - 32 \cdot - 4}}{2 \cdot 8}$

Multiply $- 32$ by $- 4$ .

$x = \frac{- 4 \pm \sqrt{16 + 128}}{2 \cdot 8}$

Add $16$ and $128$ .

$x = \frac{- 4 \pm \sqrt{144}}{2 \cdot 8}$

Rewrite $144$ as $12^{2}$ .

$x = \frac{- 4 \pm \sqrt{12^{2}}}{2 \cdot 8}$

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

$x = \frac{- 4 \pm 12}{2 \cdot 8}$

Multiply $2$ by $8$ .

$x = \frac{- 4 \pm 12}{16}$

Simplify $\frac{- 4 \pm 12}{16}$ .

$x = \frac{- 1 \pm 3}{4}$

Step 4

The final answer is the combination of both solutions.

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