Can -3 and 2 be eigenvalues of the following matrix? Can $-3$ and $2$ be eigenvalues of and nxn matrix B such that $A = B^{2}+B-6I$ and A's determinant is $0$?
So this is what I concluded:
At first glance, it can be seen that the matrix $A$ can be factored into two different terms.
\begin{align*} A = B^{2}+B-6I = (B+3I)(B-2I) \end{align*}
\begin{align*}(B+3I)(B-2I) = B^{2}-2IB+3IB-6I^{2} \end{align*}
\begin{align*} 2IB=2B, 3IB=3B, I^{2}=I.\end{align*}
\begin{align*} (B+3I)(B-2I) = B^{2}-2IB+3ID-6I^{2} =  B^{2}+B-6I = A. \end{align*}
Given that the $det(A) = 0$, let the matrix $C = B+3I$, the matrix $D = B-2I$, and 
\begin{align*}
det(A) = det(CD)=
\begin{vmatrix}CD\end{vmatrix}
=
\begin{vmatrix} C\end{vmatrix}
\begin{vmatrix} D\end{vmatrix}
= 0.
\end{align*}
In turn, this suggests that either \begin{align*} \begin{vmatrix} C \end{vmatrix}, \begin{vmatrix} D \end{vmatrix}, or (
\begin{vmatrix}C\end{vmatrix} and \begin{vmatrix}D\end{vmatrix}) = 0.\end{align*} 
So my question is... is this the correct way of proving this so far?

Your reasoning starts out fine, up to "In turn, this suggests that either
$$
|C|,|D|, \textrm{ or } |C| \textrm{ and } |D| = 0.''
$$
After that, I'm not sure what you're trying to do, but it is something circular/unnecessary.
To start over from your last correct assertion, you now know that 
$$
\det[(B + 3I)(B-2I)] = 0
$$
which tells you that either
$$
\det(B+3I) = 0
$$
or
$$
\det(B-2I) = 0
$$
(or both). If the first statement is true, then $-3$ is an eigenvalue of $B$. This is (equivalent to) the definition of eigenvalue. Similarly, if the second statement is true, then $2$ is an eigenvalue. So, at least one of those two things is true, and thus both could be eigenvalues. (But it isn't required that $B$ have both as eigenvalues, consider $B = \begin{pmatrix} 2 & 0 \\ 0 & 1\end{pmatrix}$; so the correct statement is "at least one of $2$ or $-3$ is an eigenvalue of $B$")
