+33661 You can factor it as:
$$(x-\frac{1}{2})(x-4m+\frac{1}{2})=0$$
so
$$\\ x=\frac{1}{2}\\ or\\x=4m-\frac{1}{2}$$
.
+118673 Thanks Alan,
OR
You could solve it by completing the square method
$$\\x^2-4xm+(2m-\frac{1}{4})=0\\\\
x^2-4xm\;=\;-(2m-\frac{1}{4})\\\\
x^2-4xm+(2m)^2\;=\;-2m+\frac{1}{4}+(2m)^2\\\\
(x-2m)^2\;=\;-2m+\frac{1}{4}+4m^2\\\\
(x-2m)^2\;=\;4m^2-2m+\frac{1}{4}\\\\
(x-2m)^2\;=\;4(m^2-\frac{m}{2}+\frac{1}{16})\\\\
(x-2m)^2\;=\;4(m-\frac{1}{4})^2\\\\
x-2m\;=\;\pm 2(m-\frac{1}{4})\\\\
x\;=\;2m\pm 2(m-\frac{1}{4})\\\\
x\;=\;2m+ 2m-\frac{1}{2}\qquad or \qquad x\;=\;2m- 2m+\frac{1}{2}\\\\
x\;=\;4m-\frac{1}{2}\qquad or \qquad x\;=\;\frac{1}{2}\\\\$$
