Answer
The solutions are $x=\dfrac{1}{2}\pm\dfrac{\sqrt{2}}{2}$
Work Step by Step
$4x^{2}-4x-1=0$
Take out common factor $4$ from the left side:
$4\Big(x^{2}-x-\dfrac{1}{4}\Big)=0$
Take the $4$ to divide the right side:
$x^{2}-x-\dfrac{1}{4}=\dfrac{0}{4}$
$x^{2}-x-\dfrac{1}{4}=0$
Take $\dfrac{1}{4}$ to the right side:
$x^{2}-x=\dfrac{1}{4}$
Add $\Big(\dfrac{b}{2}\Big)^{2}$ to both sides of the equation. In this case, $b=-1$
$x^{2}-x+\Big(\dfrac{-1}{2}\Big)^{2}=\dfrac{1}{4}+\Big(\dfrac{-1}{2}\Big)^{2}$
$x^{2}-x+\dfrac{1}{4}=\dfrac{1}{4}+\dfrac{1}{4}$
$x^{2}-x+\dfrac{1}{4}=\dfrac{1}{2}$
Factor the left side of the equation, which is a perfect square trinomial:
$\Big(x-\dfrac{1}{2}\Big)^{2}=\dfrac{1}{2}$
Take the square root of both sides:
$\sqrt{\Big(x-\dfrac{1}{2}\Big)^{2}}=\sqrt{\dfrac{1}{2}}$
$x-\dfrac{1}{2}=\pm\dfrac{1}{\sqrt{2}}$
$x-\dfrac{1}{2}=\pm\dfrac{\sqrt{2}}{2}$
Solve for $x$:
$x=\dfrac{1}{2}\pm\dfrac{\sqrt{2}}{2}$
