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 $- 1$ .

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

Add $16$ and $16$ .

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

Rewrite $32$ as $4^{2} \cdot 2$ .

Factor $16$ out of $32$ .

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

Rewrite $16$ as $4^{2}$ .

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

Pull terms out from under the radical.

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

Multiply $2$ by $4$ .

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

Simplify $\frac{4 \pm 4 \sqrt{2}}{8}$ .

$x = \frac{1 \pm \sqrt{2}}{2}$

Step 4

The result can be shown in multiple forms.

Exact Form:

$x = \frac{1 \pm \sqrt{2}}{2}$

Decimal Form:

$x = 1.20710678 \dots, - 0.20710678 \dots$