Answer
Characteristic equation: $\lambda^2-10\lambda+16$
$\lambda=2,8$
Work Step by Step
1.) To find the characteristic polynomial, find $det(A-\lambda I)$
$=(5-\lambda)(5-\lambda)-9$
$=\lambda^2-10\lambda+16$
2.) Solve for the eigenvalues by equating the characteristic equation to zero and solving for $\lambda$
$\lambda^2-10\lambda+16=0$
$(\lambda-2)(\lambda-8) = 0$
$\lambda=2,8$