Compare your answer:

$\begin{array}{ccc}\hfill x^2-x+\frac{3}{16}& \hfill =\hfill & 0\hfill \\ \hfill x^2-x+\frac{1}{4}& \hfill =\hfill & -\frac{3}{16}+\frac{1}{4}=-\frac{3}{16}+\frac{4}{16}=\frac{1}{16}\hfill \\ \hfill x^2-x+\frac{1}{4}& \hfill =\hfill & \frac{1}{16}\hfill \\ \hfill \sqrt{{(x-\frac{1}{2})}^2}& \hfill =\hfill & \sqrt{\frac{1}{16}}\hfill \\ \hfill x-\frac{1}{2}& \hfill =\hfill & \pm \frac{1}{4}\hfill \\ \hfill x-\frac{1}{2}& \hfill =\hfill & \frac{1}{4}\hfill \\ \hfill x-\frac{1}{2}& \hfill =\hfill & -\frac{1}{4}\hfill \end{array}$

$x=\frac{1}{4}\text{ and }x=\frac{3}{4}$