When is $\mathbb{Q}[x]/f(x) \cong \mathbb{Q}[x]/g(x)$, if $f(x),g(x) \in \mathbb{Q}$ are polynomials of degree 2?
My initial approach to this question was to note that $\mathbb{Q}[x]/f(x) \cong \mathbb{Q}[x]/g(x)$ iff $\mathbb{Q}(\sqrt{a}) \cong \mathbb{Q}(\sqrt{b})$ for roots $\sqrt{a}$ of $f(x)$ and $\sqrt{b}$ of $g(x)$, but I'm unsure if this is actually a fruitful approach.
The isomorphism class is determined by the discriminant of $f(x)$, in case of degree 2. If $f(x)=x^{2} + ax + b$ and $g(x) = x^{2} + cx + d$ are both irreducible, then $\mathbb{Q}[x]/(f(x)) \simeq \mathbb{Q}(\sqrt{a^{2}- 4b})$, and two fields are isomorphic if and only if $(a^{2} - 4b)/(c^{2} - 4d)$ is a square of some rational number. If part is not hard to show, and only if part is basically same as the fact that $\mathbb{Q}(\sqrt{2})$ and $\mathbb{Q}(\sqrt{3})$ are not isomorphic as fields.