Answer
$x=0$ and $x=\dfrac{1}{4}$
Work Step by Step
$4x^{2}-x=0$
First, take out common factor $4$:
$4\Big(x^{2}-\dfrac{1}{4}x\Big)=0$
Take the $4$ to divide the right side of the equation. We get:
$x^{2}-\dfrac{1}{4}x=0$
Add $\Big(\dfrac{b}{2}\Big)^{2}$ to both sides of the equation. For this particular problem, $b=-\dfrac{1}{4}$
$x^{2}-\dfrac{1}{4}x+\Big(\dfrac{-1/4}{2}\Big)^{2}=\Big(\dfrac{-1/4}{2}\Big)^{2}$
$x^{2}-\dfrac{1}{4}x+\dfrac{1}{64}=\dfrac{1}{64}$
Factor the perfect square trinomial that we have on the left side of the equation:
$\Big(x-\dfrac{1}{8}\Big)^{2}=\dfrac{1}{64}$
Solve for $x$:
$x-\dfrac{1}{8}=\pm\sqrt{\dfrac{1}{64}}$
$x=\dfrac{1}{8}\pm\dfrac{1}{8}$
So, our two solutions are:
$x=\dfrac{1}{8}-\dfrac{1}{8}=0$ and $x=\dfrac{1}{8}+\dfrac{1}{8}=\dfrac{1}{4}$
