Answer
$\displaystyle \frac{1}{4}\pm\frac{1}{4}i$
Work Step by Step
$8x^{2}-4x+1=0$
We solve using the quadratic formula ($a=8,\ b=-4,\ c=1$):
$\displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
$\displaystyle x=\frac{-(-4)\pm\sqrt{(-4)^{2}-4(8)(1)}}{2(8)}$
$\displaystyle x=\frac{-(-4)\pm\sqrt{16-32}}{2(8)}$
$\displaystyle x=\frac{+4\pm\sqrt{-16}}{16}$
$\displaystyle x=\frac{4\pm 4i}{16}=\frac{1}{4}\pm\frac{1}{4}i$
Recall, $i=\sqrt{-1}$
