Let $$A=\begin{pmatrix}a&\!\!b\\c&d\end{pmatrix}$$ be a $2\times2$ matrix with real entries. After some computations, the characteristic polynomial of $A$ is
$$\lambda^2-(a+d)\lambda + (ad - bc).$$
Finding its zeroes by the quadratic formula we get
$$\lambda = \frac{(a+d) \pm \sqrt{(a+d)^{2}-4(ad-bc)}}{2} = \frac{(a+d) \pm \sqrt{(a-d)^{2}-4bc}}{2}$$
I know for a fact that if $(a-d)^{2}-4bc > 0$, $A$ would be diagonalizable since we have two distinct real eigenvalues. Now i wonder what happens when $(a-d)^{2}-4bc < 0$. I'm sure that the eigenvalues are complex if that's the case but I'm unsure if it is diagonalizable.
Even if $(a-d)^{2}-4bc = 0$, the matrix can be diagonalizable. The identity matrix provides an example.
If the discriminant is negative, then the eigenvalues are not real and the matrix can't be diagonalizable on $\mathbb R$. It can be on $\mathbb C$ though.