Answer
$\displaystyle \frac{1}{2}\pm\frac{\sqrt{23}}{2}i$
Work Step by Step
$x(1-x)=6$
$x-x^2=6$
$-x^{2}+x-6=0$
We solve using the quadratic formula ($a=-1,\ b=1,\ c=-6$):
$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
$x=\frac{-1\pm\sqrt{1^{2}-4*-1*-6}}{2(-1)}$
$x=\frac{-1\pm\sqrt{1-24}}{2(-1)}$
$x=\frac{-1\pm\sqrt{-23}}{-2}$
$x=\frac{-1\pm\sqrt{23}i}{-2}$
$x=\frac{1}{2}\pm\frac{\sqrt{23}}{2}i$
