How do you quickly know that this matrix is diagonalisable (characteristic polynomial given)? We have matrix $A=\begin{pmatrix}
3 & 4 & -3\\ 
2 & 7 & -4\\ 
3 & 9 & -5
\end{pmatrix}$
The characteristic polynomial is $-(\lambda-2)^{2} \cdot (\lambda-1)=0$
Now I'd like to know a quick way to know if this matrix is diagonalisable. To be more precise, I want know that this matrix has as many eigenvalues as its own size (it's a $3 \times 3$ matrix so we need at least$3$ eigenvalues).
We see that $\lambda_{1}=2$ and $\lambda_{2}=1$ are eigenvalues. But how do I know without further long calculation that $\lambda_{1}=2$ is a double eigenvalue? I only know it because I did further calculations (polynomial long division) but how can I know it without wasting more time?

I think that there's a simple test that requires really little further computation. If you want to check if $\lambda$ is a double eigenvalue of $A$, then it is sufficient to see if $A-\lambda I$ has rank $n-2$.
In your case, if $A-2I$ has rank 1, and rank 1 matrices are really simple, since all the columns are multiples of each other.
$$
A - 2I=\begin{pmatrix}
1 & 4 & -3\\ 
2 & 5 & -4\\ 
3 & 9 & -7
\end{pmatrix}
$$
In this case,  2 isn't a double eigenvalue, since the first and second column are not multiple of each other, so this matrix has rank 2 or more.
