# What are the eigenvalues of this symmetric matrix?

Let

$$A=\begin{pmatrix} 3 & 2 & 4 \\ 2 & 0 & 2 \\ 4 & 2 & 3 \end{pmatrix}.$$

I'm trying to find the eigenvalues of $A$, but when I calculate the characteristic polynomial, I get $$p(\lambda)=-\lambda^3+6\lambda^2+15\lambda+2,$$ and I don't know how to solve $p(\lambda)=0$. I'd appreciate any help. Thanks in advance.

$$p(\lambda)=-\lambda^3+6\lambda^2+15\lambda+8 = -(\lambda-8)(\lambda+1)^2 = 0$$
So you have the two eigenvalues $\lambda_1 = 8$ (single root) and $\lambda_{2,3} = -1$ (double root).