Minimal polynomial of $\sqrt{3+2\sqrt{3}}$ Let $a=\sqrt{3+2\sqrt{3}}$. Then
\begin{align*}
&a=\sqrt{3+2\sqrt{3}}\\
&\implies a^2=3+2\sqrt{3}\\
&\implies a^2-3 = 2\sqrt{3}\\
&\implies (a^2-3)^2 = 4\cdot 3=12\\
&\implies (a^2-3)^2-12=0.
\end{align*}
So, $a$ a root of
$$(x^2-3)^2-12=x^4-6x^2-3.$$
By Eisenstein's criterion, $x^4-6x^2-3$ is irreducible over $\mathbb Q$. So, $a$ is algebraic of degree $4$.

However, I've seen that the degree of $\mathbb Q(\sqrt{3+2\sqrt{3}})$ over $\mathbb Q$ is $2$.
What is wrong with my attempt?

You are unquestionably right. A much more advanced viewpoint:
The ring $R=\Bbb Z[\sqrt3\,]$ is the integer-ring of the quadratic number field $K=\Bbb Q(\sqrt3\,)$. It’s “well known” (and easy to prove) that $R$ is a Principal Ideal Domain (class number is $1$).
We have the factorization $z=3+2\sqrt3=\sqrt3(2+\sqrt3\,)$, in which the two factors are respectively, an indecomposable element, and a generator of the (free part of the) unit group. In particular, neither is a square, and they are independent modulo squares. Therefore, the square root $\sqrt z$ generates a quadratic extension of $K$. That does it.
