\(2x^2-2ix+10=0\implies x=\dfrac{-(-2i)\pm\sqrt{(-2i)^2-4\times2\times10}}{2\times2}=\dfrac{2i\pm\sqrt{-84}}4\)
When you add the roots, the square root terms will cancel, leaving you with
\(S=\dfrac{2i}4+\dfrac{2i}4=\dfrac{4i}4=i\)
and so \(S^{10}=i^{10}=(i^4)^2i^2=-1\).
