Answer
There are no real solutions.
Work Step by Step
We solve:
$x(1-x)=6$
$x-x^{2}=6$
$x^{2}-x+6=0$
We solve using the quadratic formula ($a=1,\ b=-1,\ c=6$):
$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
However, we note that the discriminant is negative:
$b^{2}-4ac=(-1)^{2}-4*1*6=1-24=-23$
Therefore, there are no real solutions.