Answer
$-\frac{1}{4}\pm\frac{\sqrt{3}}{4}i$
Work Step by Step
$z^{2}+\displaystyle \frac{1}{2}z+\frac{1}{4}=0$
First we multiply through by 4:
$4z^{2}+2z+1=0 $
Next, we solve using the quadratic formula (a=4, b=2, c=1):
$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
$z=\displaystyle \frac{-2\pm\sqrt{2^{2}-4(4)(1)}}{2(4)}$
$=\frac{-2\pm\sqrt{4-16}}{8}$
$=\frac{-2\pm\sqrt{-12}}{8}$
$=\frac{-2\pm\sqrt{-1*3*4}}{8}$
We use the fact that $\sqrt{-1}=i$:
$=\frac{-2\pm 2\sqrt{3}i}{8}$
$=-\frac{1}{4}\pm\frac{\sqrt{3}}{4}i$
