Prove that $\mathbb{Q}[x]/\langle 3+3x+x^2\rangle \cong \mathbb{Q}(i\sqrt{3})$

Question

If $\mathbb{Q}(i\sqrt{3})=\left\{a+b(i\sqrt{3}) \:\vert\: a,b \in\mathbb{Q}\right\}$ then $\mathbb{Q}[x]/\langle 3+3x+x^2\rangle \cong \mathbb{Q}(i\sqrt{3})$. An earlier part of this problem asks to prove that if $\lambda$ is a root of a polynomial $a_0+b_0x+x^2$ then $\mathbb{Q}(\lambda=\left\{a+b\lambda)\:\vert\: a,b \in\mathbb{Q}\right\}$ is a subfield of $\mathbb{C}$ and I've proved it. However, I'm not sure if it can help me solve this question.