In general, $F(\alpha)/F$ is Galois if and only if the minimal polynomial of $\alpha$ over $F$ is separable and splits completely in $F(\alpha)$. In the special case of quadratic extensions, it is very easy to see that the "splits completely" part is always satisfied, so it's just a question of separability. If for example $F$ has characteristic 0, then separability is automatic, so in that case, any quadratic extension is Galois.
There do exist Galois extensions of $\mathbb{Q}$ of degree 3. If $\alpha$ has cubic minimal polynomial over $\mathbb{Q}$, then $\mathbb{Q}(\alpha)$ is Galois if and only if the discriminant of this polynomial is a rational square. This is proved in any standard text on Galois theory.