Calculate eigenvalues and eigenvectors

You can stop at the step $$(2-\lambda)(\lambda^2-4\lambda + 5) = 0.$$

Here you get the roots $\lambda = 2$ or $\lambda = 2 \pm i$.

It happens that, if $\lambda=2$, then $-\lambda^3+6\lambda^2-13\lambda+10=0$. In fact, $-\lambda^3+6\lambda^2-13\lambda+10=(2-\lambda)(\lambda^2-4\lambda+5)$. The roots of the polynomial $\lambda^2-4\lambda+5$ are complex non-real numbers: $2+i$ and $2-i$.