Eigenvalues of matrix $A$ $n \times n$, knowing that $A^2+A-6I_n=O_n$ How can I find ALL eigenvalues of matrix $A$, $n \times n$, $n \ge 3$, knowing that 
$$A^2+A-6I_n=O_n.$$
I applied the Cayley-Hamilton theorem, and obviously two of the eigenvalues are $-3$ and $2$.
However, how do I find the rest of them?

Let's try to replicate the minimal polynomial theorem, though in this problem, nothing says that the given polynomial is minimal.
Let $\lambda$ be an eigenvalue of $A$ and $v$ an eigenvector of $\lambda$. Then 
$$ A^i v = \lambda ^i v, \text{ for } i=0,1,2,\dots$$
Thus
$$\lambda^2 + \lambda -6 =0.$$
Thus the only two possible eigenvalues are $-3$ and $2$. On the other hand, since
$$(A+3I)(A-2I) = 0,$$
So if $\mathop{\mathrm{Im}}(A-2I) \neq 0$, that is $A\neq 2I$, then $\mathrm{Ker}(A+3I)\neq 0$ or $-3$ is an eigenvalue. Similarly argument for eigenvalue $2$.
