What what value of a is A diagonalizable? Let $$A=\begin{pmatrix}3&a\\ 2&0\end{pmatrix}.$$
For what value of a is the diagonal factorization $A=SΛS^{-1}$ impossible?
Since it's not a triangular matrix, I cannot read the eigenvalues off the diagonal directly. I've found out that the polynomial is $λ^2-3λ-2a$. Solving for λ, I get $λ=\frac{3}{2}\pm \sqrt{2a+\frac{9}{4}}$
Not sure how to go after this.
 A: For a matrix to fail to be diagonalizable, you need that at least one eigenvalue has multiplicity two, and that the dimension of the eigenspace is strictly less than the multiplicity. 
Here since you know the characteristic polynomial, you can see that there will be single eigenvalue when $a=-9/8$, and the eigenvalue is $3/2$. Now let us check if $A$ is diagonalizable in that case: an eigenvector for $3/2$ will be a solution to 
$$
\begin{bmatrix} 3-3/2&-9/8\\ 2&-3/2\end{bmatrix} \begin{bmatrix} x\\ y\end{bmatrix} =\begin{bmatrix} 0\\0\end{bmatrix} ,
$$
which gives $12x=9y$. So all eigenvectors for $3/2$ are of the form $t\begin{bmatrix} 9\\12\end{bmatrix}$. That is, the eigenspace of $3/2$ is one-dimensional, while its multiplicity is two, so $A$ is not diagonalizable. 
The above computation can be avoided is one notices that the only way a $2\times2$ matrix can have an eigenvalue $\lambda$ with eigenspace of dimension $2$ is if $A=\lambda I$. As your matrix is not diagonal, one can deduce that it is not diagonalizable without computing the eigenvectors. 
Edit: as mentioned by amd, if we restrict the field to $\mathbb R$, then there are no eigenvalues when $a<9/8$, thus making $A$ not diagonalizable in that context. 
