Prove $\displaystyle \frac{\mathbb{Q}[x]}{<x^2+4x+2>} \cong \frac{\mathbb{Q}[x]}{<x^2-2>} $. It's obvious that $\displaystyle \frac{\mathbb{Q}[x]}{<x^2-2>} \cong \mathbb{Q}[\sqrt{2}]$ but I can't get the other way, and it seems to be a logical way to prove this.

Any help is appreciated.