$\displaystyle $(2x-1)^{2}=-4$$
I thought this could only be used if k (the constant) was positive.
You are right, this does not have any real solutions.
It does have two complex solutions though...
$\displaystyle \displaystyle (2x - 1)^2 = -4$
$\displaystyle \displaystyle 2x - 1 = \pm \sqrt{-4}$
$\displaystyle \displaystyle 2x - 1 = \pm 2i$
$\displaystyle \displaystyle 2x = 1 \pm 2i$
$\displaystyle \displaystyle x = \frac{1}{2} \pm i$.