Find All Complex Solutions 0=16x^2+8x-3

$0 = 16 x^{2} + 8 x - 3$

Step 1

Move all the expressions to the left side of the equation.

Subtract $16 x^{2}$ from both sides of the equation.

$- 16 x^{2} = 8 x - 3$

Subtract $8 x$ from both sides of the equation.

$- 16 x^{2} - 8 x = - 3$

Add $3$ to both sides of the equation.

$- 16 x^{2} - 8 x + 3 = 0$

Step 2

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 3

Substitute the values $a = - 16$ , $b = - 8$ , and $c = 3$ into the quadratic formula and solve for $x$ .

$\frac{8 \pm \sqrt{{(- 8)}^{2} - 4 \cdot (- 16 \cdot 3)}}{2 \cdot - 16}$

Step 4

Simplify.

Simplify the numerator.

Raise $- 8$ to the power of $2$ .

$x = \frac{8 \pm \sqrt{64 - 4 \cdot - 16 \cdot 3}}{2 \cdot - 16}$

Multiply $- 4 \cdot - 16 \cdot 3$ .

Multiply $- 4$ by $- 16$ .

$x = \frac{8 \pm \sqrt{64 + 64 \cdot 3}}{2 \cdot - 16}$

Multiply $64$ by $3$ .

$x = \frac{8 \pm \sqrt{64 + 192}}{2 \cdot - 16}$

Add $64$ and $192$ .

$x = \frac{8 \pm \sqrt{256}}{2 \cdot - 16}$

Rewrite $256$ as $16^{2}$ .

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

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

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

Multiply $2$ by $- 16$ .

$x = \frac{8 \pm 16}{- 32}$

Simplify $\frac{8 \pm 16}{- 32}$ .

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

Move the negative in front of the fraction.

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

Step 5

The final answer is the combination of both solutions.

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