Let $i=\sqrt{-1}$. The product of the real parts of the roots of $z^2-z=5-5i$ is
$\text{(A) } -25\quad \text{(B) } -6\quad \text{(C) } -5\quad \text{(D) } \frac{1}{4}\quad \text{(E) } 25$
Applying the quadratic formula gives \[z=\frac{-1\pm\sqrt{21-20i}}{2}\]
Let \[\sqrt{21-20i}=a+bi\] where $a$ and $b$ are real. Squaring both sides and equating real and imaginary parts gives \[a^2-b^2=21\] \[2ab=-20\] Substituting $b=-\frac{10}{a}$, letting $a^2=n$ and solving gives $n=25, -4$, from which we see that $a=5$ and $b=-2$.
Replacing the square root in the quadratic formula with $5-2i$ and simplifying gives the two roots $2-i$ and $-3+i$. The product of their real parts is $-6$. The answer is $\fbox{B}$.