Answer
$x=\dfrac{1}{4}$ and $x=\dfrac{1}{2}$
Work Step by Step
$x^{2}=\dfrac{3}{4}x-\dfrac{1}{8}$
Take $\dfrac{3}{4}x$ substracting to the left side:
$x^{2}-\dfrac{3}{4}x=-\dfrac{1}{8}$
Add $\Big(\dfrac{b}{2}\Big)^{2}$ to both sides of the equation. For this particular problem, $b=-\dfrac{3}{4}$:
$x^{2}-\dfrac{3}{4}x+\Big(\dfrac{-3/4}{2}\Big)^{2}=-\dfrac{1}{8}+\Big(\dfrac{-3/4}{2}\Big)^{2}$
$x^{2}-\dfrac{3}{4}x+\dfrac{9}{64}=-\dfrac{1}{8}+\dfrac{9}{64}$
$x^{2}-\dfrac{3}{4}x+\dfrac{9}{64}=\dfrac{1}{64}$
Factor the perfect square trinomial we have on the left side:
$\Big(x-\dfrac{3}{8}\Big)^{2}=\dfrac{1}{64}$
Solve for $x$:
$x-\dfrac{3}{8}=\pm\sqrt{\dfrac{1}{64}}$
$x-\dfrac{3}{8}=\pm\dfrac{1}{8}$
$x=\dfrac{3}{8}\pm\dfrac{1}{8}$
So, our two solutions are:
$x=\dfrac{3}{8}-\dfrac{1}{8}=\dfrac{1}{4}$ and $x=\dfrac{3}{8}+\dfrac{1}{8}=\dfrac{1}{2}$
