A much quicker, albeit high teck, solution is as follows. We know that there exists a unique ring map $\displaystyle \mathbb{Q}[x]\to\mathbb{Q}[x]$ determined by $\displaystyle x\mapsto x+2$ (it's just $\displaystyle p(x)\mapsto p(x+2)$) call this ring map $\displaystyle f$. Note then that we can compose this with the canonical projection $\displaystyle \mathbb{Q}[x]\to\mathbb{Q}[x]/(x^2+4x+2)$ to get a ring map $\displaystyle \overline{f}:\mathbb{Q}[x]\to\mathbb{Q}[x]/(x^2+4x+2)$. Now, note that $\displaystyle f(x^2-2)=(x+2)^2-2=x^2+4x+2$ so that $\displaystyle (x^2-2)\subseteq\ker\overline{f}$. So, $\displaystyle \overline{f}$ descends to a well-defined ring map $\displaystyle g:\mathbb{Q}[x]/(x^2-2)\to\mathbb{Q}[x]/(x^2+4x+2)$. Now, note that this ring map is automatically injective since $\displaystyle \mathbb{Q}[x]/(x^2-2)$ is a field (clearly $\displaystyle x^2-2$ is irreducible) and that this ring map is surjective because $\displaystyle g$ is a $\displaystyle \mathbb{Q}$-algebra map and $\displaystyle \mathbb{Q}[x]/(x^2-2)$ and $\displaystyle \mathbb{Q}[x]/(x^2+4x+2)$ are both $\displaystyle 2$-dimensional $\displaystyle \mathbb{Q}$-spaces. So, $\displaystyle g$ is an isomorphism.
To reiterate, Deveno gave the better intuition for why this is true. Namely, $\displaystyle \mathbb{Q}[x]/(x^2-2)$ is the "splitting field" for $\displaystyle x^2-2$ and similarly $\displaystyle \mathbb{Q}[x]/(x^2+4x+2)$ is the splitting field for $\displaystyle x^2+4x+2$. Now, any two splitting fields are isomorphic, and you can show that $\displaystyle \mathbb{Q}(\sqrt{2})$ (literally the subfield of $\displaystyle \mathbb{R}$ generated by $\displaystyle \mathbb{Q}$ and $\displaystyle \sqrt{2}$) is also a splitting field for $\displaystyle x^2-2$ (over $\displaystyle \mathbb{Q}$) so that $\displaystyle \mathbb{Q}[x]/(x^2-2)\cong\mathbb{Q}(\sqrt{2})\subseteq\mathbb{R}$. But, using the same logic, one can prove that $\displaystyle \mathbb{Q}(\sqrt{2}-2)$ is a splitting field for $\displaystyle \mathbb{Q}[x]/(x^2+4x+2)$ and so $\displaystyle \mathbb{Q}[x]/(x^2+4x+2)\cong\mathbb{Q}(\sqrt{2}-2)\subseteq\mathbb{R}$. Since, by inspection, $\displaystyle \mathbb{Q}(\sqrt{2})=\mathbb{Q}(\sqrt{2}-2)$ (since you are already including $\displaystyle \mathbb{Q}$ in the generating set, there is no distinction between $\displaystyle \sqrt{2}$ and $\displaystyle \sqrt{2}-2$) you can see that $\displaystyle \mathbb{Q}[x]/(x^2-2)\cong\mathbb{Q}[x]/(x^2+4x+2)$.
Note: The above actually goes to show that if [tex/]p(x)[/tex] and $\displaystyle q(x)$ are rational (more generally of course) polynomials such that $\displaystyle p(x+a)=q(x)$ for some rational $\displaystyle a$, then $\displaystyle \mathbb{Q}[x]/(p(x))\cong\mathbb{Q}[x]/(q(x))$--the same intuition though applies that
$\displaystyle \begin{aligned}\mathbb{Q}[x]/(p(x)) &\cong\mathbb{Q}(\text{roots of p})\\ &=\mathbb{Q}(\text{rationally perturbed roots of }q)\\ &=\mathbb{Q}(\text{roots of }q)\\ &\cong\mathbb{Q}[x]/(q(x))\end{aligned}$.