Quadratic formula with a=1,b=0,c=-25 gives lambda = plus-or-minus 5. Verify with eigenvector for lambda=5: solve (A-5I)v=0, giving v=(1,2). Check Av = (5,10) = 5v, confirming lambda=5 is genuine; the other eigenvalue is -5 since eigenvalues sum to the trace, 0.
\[ \lambda = -5,\ 5 \Rightarrow \text{Option (B)} \]