Solve Using the Quadratic Formula -4x^2-4x-1=0

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

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

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

Step 3

Simplify.

Simplify the numerator.

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

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

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

Multiply $- 4$ by $- 4$ .

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

Multiply $16$ by $- 1$ .

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

Subtract $16$ from $16$ .

$x = \frac{4 \pm \sqrt{0}}{2 \cdot - 4}$

Rewrite $0$ as $0^{2}$ .

$x = \frac{4 \pm \sqrt{0^{2}}}{2 \cdot - 4}$

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

$x = \frac{4 \pm 0}{2 \cdot - 4}$

$4$ plus or minus $0$ is $4$ .

$x = \frac{4}{2 \cdot - 4}$

Multiply $2$ by $- 4$ .

$x = \frac{4}{- 8}$

Cancel the common factor of $4$ and $- 8$ .

Factor $4$ out of $4$ .

$x = \frac{4 (1)}{- 8}$

Cancel the common factors.

Factor $4$ out of $- 8$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Move the negative in front of the fraction.

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

Step 4

The final answer is the combination of both solutions.

$x = - \frac{1}{2}$ Double roots