Find All Complex Solutions -4x^2-2x-1=0

$- 4 x^{2} - 2 x - 1 = 0$

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 $a = - 4$ , $b = - 2$ , and $c = - 1$ into the quadratic formula and solve for $x$ .

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

Step 3

Simplify.

Simplify the numerator.

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

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

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

Multiply $- 4$ by $- 4$ .

$x = \frac{2 \pm \sqrt{4 + 16 \cdot - 1}}{2 \cdot - 4}$

Multiply $16$ by $- 1$ .

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

Subtract $16$ from $4$ .

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

Rewrite $- 12$ as $- 1 (12)$ .

$x = \frac{2 \pm \sqrt{- 1 \cdot 12}}{2 \cdot - 4}$

Rewrite $\sqrt{- 1 (12)}$ as $\sqrt{- 1} \cdot \sqrt{12}$ .

$x = \frac{2 \pm \sqrt{- 1} \cdot \sqrt{12}}{2 \cdot - 4}$

Rewrite $\sqrt{- 1}$ as $i$ .

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

Rewrite $12$ as $2^{2} \cdot 3$ .

Factor $4$ out of $12$ .

$x = \frac{2 \pm i \cdot \sqrt{4 (3)}}{2 \cdot - 4}$

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

$x = \frac{2 \pm i \cdot \sqrt{2^{2} \cdot 3}}{2 \cdot - 4}$

Pull terms out from under the radical.

$x = \frac{2 \pm i \cdot (2 \sqrt{3})}{2 \cdot - 4}$

Move $2$ to the left of $i$ .

$x = \frac{2 \pm 2 i \sqrt{3}}{2 \cdot - 4}$

Multiply $2$ by $- 4$ .

$x = \frac{2 \pm 2 i \sqrt{3}}{- 8}$

Simplify $\frac{2 \pm 2 i \sqrt{3}}{- 8}$ .

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

Move the negative in front of the fraction.

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