For which $x$ is $$M=\begin{pmatrix}4&0&-2\\x&5&4\\0&0&5\end{pmatrix}$$ diagonalizable?
I know a matrix which is diagonalizable can be written in the form $A=S\Lambda S^{-1}$ But I don't know what property these kinds of matrices all have? Distinct eigenvalues?
For your given matrix it is easy to see, that the eigenvalues of $M$ are $4,5$ and $5$, and this holds independent of $x$
Thus $M$ is diagonalizable if $\operatorname{dim}\operatorname{Eig}(5) = 2$. To calculate the eigenspace to $5$ you have to solve $A\nu =5\nu$ for $\nu$, which is equivalent to
$$A\nu=5\nu\Longleftrightarrow \begin{pmatrix}-1&0&-2\\x&0&4\\0&0&0\end{pmatrix}\cdot \nu = \begin{pmatrix}0\\0\\0\end{pmatrix}$$
One eigenvector $\nu_1$ is clearly given by $\left(0,1,0\right)^T$. A second linear unindependent eigenvector $\nu_2$ is given by $\left(2,0,-1\right)^T$ iff $x=2$.