The eigenvalues are the solutions to $\det(A-\lambda I)=0$. Since $A-\lambda I=\begin{bmatrix}2-\lambda&-5\\4&-2-\lambda\end{bmatrix}\;,$ $\det(A-\lambda I)=(2-\lambda)(-2-\lambda)+20=\lambda^2+16$, and $\lambda=\pm4i$. You should now take each of these eigenvalues and solve the system $A\vec x=\lambda\vec x$, i.e., $(A-\lambda I)\vec x=\vec 0$. For example, with $\lambda=4i$ you have to solve
$\begin{bmatrix}2-4i&5\\4&-2-4i\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}\;.$
It’s not as messy as it looks: just multiply the first row by $2$ and the second by $1-2i$, and you get the augmented matrix
$\begin{bmatrix}4-8i&-10&0\\4-8i&-10&0\end{bmatrix}\;,$
since $(1-2i)(-2-4i)=-2(1-2i)(1+2i)=-2(1-4i^2)=-10$.
Added: Since it may be the manipulations with complex numbers that are causing the difficulty, I’ll go ahead and finish the reduction. The single non-zero row reduces to $\begin{bmatrix}2-4i&-5&0\end{bmatrix}$, so to finish the reduction you need to calculate $\frac{-5}{2-4i}=\frac{-5}{2-4i}\cdot\frac{2+4i}{2+4i}=\frac{-10-20i}{4-16i^2}=\frac{-10-20i}{20}=-\frac12-i\;.$ This gives you the eigenvector $\begin{bmatrix}\frac12+i\\1\end{bmatrix}\;,$ and indeed
$\begin{bmatrix}2&-5\\4&-2\end{bmatrix}\begin{bmatrix}\frac12+i\\1\end{bmatrix}=\begin{bmatrix}-4+2i\\4i\end{bmatrix}=4i\begin{bmatrix}\frac12+i\\1\end{bmatrix}\;.$