Diagonalization & Matrix Powers · Hard problem

Let $\mathbf{x}(t)$ solve $\mathbf{x}' = A\mathbf{x}$ with $A = \begin{pmatrix}1 & 2\\2 & 1\end{pmatrix}$ and $\mathbf{x}(0) = \begin{pmatrix}3\\-1\end{pmatrix}$. The second component $x_2(t)$ starts negative and becomes $0$ at exactly one time $t > 0$. Find that time.