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We can write $z^2-z-(5-5i)=0.$ By the quadratic formula, $$z=\frac{1\pm \sqrt{1+4(5-5i)}}{2}=\frac{1\pm \sqrt{21-20i}}{2}.$$Let $21-20i=(a+bi{)}^2,$ where $a$ and $b$ are real numbers. This expands as $$a^2+2abi-b^2=21-20i.$$Equating the real and imaginary parts, we get $a^2-b^2=21$ and $ab=-10,$ so $b=-\frac{10}{a}.$ Substituting, we get $$a^2-\frac{100}{a^2}=21.$$Then $a^4-21a^2-100=0,$ which factors as $(a^2-25)(a^2+4)=0.$ Since $a$ is real, $a^2=25,$ which means $a=5$ or $a=-5.$

If $a=5,$ then $b=-2,$ so $$z=\frac{1+5-2i}{2}=3-i.$$If $a=-5,$ then $b=2,$ so $$z=\frac{1-5+2i}{2}=-2+i.$$Therefore, the solutions are $\boxed{3-i,-2+i}.$