Suppose $A\in M_n(C)$ and define $B = \begin{pmatrix}O & A\\ A & 2A\end{pmatrix}$. Find eigenvalues of $B$ in terms of $A$.
I know that to find eigenvalues of $B$ we have to solve $det(B-xI)=0$. Also, $B$ has square blocks of the same order so based on Determinant of Block matrices: $det \begin{pmatrix}-xI & A\\A & 2A-xI\end{pmatrix}= det((-xI)(2A-xI) -A^2)$.
But I don't know how to solve the equation.
On
Let $(V_k,\lambda_k$) be the eigenpairs of $A$, meaning that:
$$AV_k=\lambda_kV_k\tag{1}$$
Let us look for eigenvectors of $B$ under the following form:
$$\begin{pmatrix}O & A\\ A & 2A\end{pmatrix}\begin{pmatrix}V_k\\ aV_k\end{pmatrix}=\mu_k\begin{pmatrix}V_k\\ aV_k\end{pmatrix}$$
which is equivalent (due to (1)) to the fact that:
$$\begin{cases} a\lambda_kV_k&=&\mu_kV_k\\ (1+2a)\lambda_kV_k&=&a\mu_kV_k\end{cases} \ \implies \ \begin{cases} a\lambda &=&\mu_k\\ (1+2a)\lambda_k&=&a\mu_k\end{cases}\tag{2}$$
therefore, we have the proportionnality:
$$\dfrac{1+2a}{a}=a$$
a quadratic equation with roots $a'=1-\sqrt{2}$ and $a''=1+\sqrt{2}$.
Therefore, using the fact that $a\lambda_k=\mu_k$, the $a'\lambda_k$ and the $a''\lambda_k$ are eigenvalues with specific eigenvectors.
As $B$ is a $2n \times 2n$ matrix, we have thus found all of them.
Remark: The idea to look for eigenvectors having the given form comes from the habit to work on the Kronecker form as in the solution given by @Ben Grossmann.
The quick approach is to simply note that $B$ is the Kronecker product $B = M \otimes A$, where $$ M = \pmatrix{0&1\\1&2}. $$ Because the eigenvalues of $M$ are $1 \pm \sqrt{2}$, the properties of the Kronecker product imply that the eigenvalues of $B$ will be of the form $\lambda (1 \pm \sqrt{2})$ for all eigenvalues $\lambda$ of $A$.
Alternatively, we can continue your approach to note that $$ \det(M - xI) = \det(-xI(2A-xI) -A^2) = \det(x^2 I - 2x A - A^2)\\ = \det([xI - (1 - \sqrt{2})A][xI - (1 + \sqrt{2})A]). $$