Proof of the ring isomorphism: $ \mathbb{Z}[\sqrt{7}] /(5+2 \sqrt{7}) \cong \mathbb{Z} / (3) $ 
I am asked to prove that 
  $$
\mathbb{Z}[\sqrt{7}] /(5+2 \sqrt{7}) \cong \mathbb{Z} / (3)
$$

First, if we define the following homorphism : $\phi:\mathbb{Z} \rightarrow \mathbb{Z}[\sqrt{7}] /(5+2 \sqrt{7})$, we can try to show that this homomorphism has $\ker(\phi)=(3)$ and is surjective to conclude with the first isomorphism theorem.
It is kind of informal but I realised that in $(5+2 \sqrt{7})$, $$5+2 \sqrt{7} \equiv 0$$
$$5 \equiv  -2 \sqrt{7}$$
$$25\equiv 28$$
$$3 \equiv 0$$
But then I struggle to show properly that $\phi (3) = 0$. 
 A: Consider the natural homomorphism $$\varphi \colon \mathbb{Z} \hookrightarrow \mathbb{Z}\left[ \sqrt{7} \right] \twoheadrightarrow \frac{\mathbb{Z}\left[ \sqrt{7} \right]}{\left\langle 5 +2 \sqrt{7} \right\rangle}$$ which sends any integer $n \in \mathbb{Z}$ to its class $n +\left\langle 5+ 2 \sqrt{7} \right\rangle \in \frac{\mathbb{Z}\left[ \sqrt{7} \right]}{\left\langle 5 +2 \sqrt{7} \right\rangle}$.
As you said, it is enough to prove that $\text{Ker}(\varphi) = 3 \mathbb{Z}$ and $\varphi$ is surjective.
We have $3 = -\left( 5 -2 \sqrt{7} \right) \left( 5 +2 \sqrt{7} \right) \in \left\langle 5 +2 \sqrt{7} \right\rangle$, and hence $3 \mathbb{Z} \subset \text{Ker}(\varphi)$. Therefore, either $\text{Ker}(\varphi) = \mathbb{Z}$ or $\text{Ker}(\varphi) = 3 \mathbb{Z}$.
Now, if we had $\text{Ker}(\varphi) = \mathbb{Z}$, there would exist $a, b \in \mathbb{Z}$ such that $1 = \left( a +b \sqrt{7} \right) \left( 5 +2 \sqrt{7} \right)$, which would yield $5 a +14 b = 1$ and $2 a +5 b = 0$ since $\sqrt{7}$ is irrational. This is impossible. Thus, $\text{Ker}(\varphi) = 3 \mathbb{Z}$.
Finally, note that $\sqrt{7} = 2 +\left( 8 -3 \sqrt{7} \right) \left( 5 +2 \sqrt{7} \right)$. Therefore, for all $a, b \in \mathbb{Z}$, we have $$a +b \sqrt{7} +\left\langle 5 +2 \sqrt{7} \right\rangle = \varphi(a +2 b) \, \text{,}$$ which proves that $\varphi$ is surjective.
P.S.: In order to avoid possible confusions, I denote by $\left\langle 5 +2 \sqrt{7} \right\rangle$ the ideal of $\mathbb{Z}\left[ \sqrt{7} \right]$ generated by $5 +2 \sqrt{7}$.
A: You're on the right path. However, I'd observe that
$$
(5+2\sqrt{7})(5-2\sqrt{7})=25-28=-3
$$
which implies that $3\in\langle5+2\sqrt{7}\rangle$. Thus the unique ring homomorphism $\varphi\colon\mathbb{Z}\to\mathbb{Z}[\sqrt{7}]/\langle5+2\sqrt{7}\rangle$ defined by $\varphi(n)=n+\langle5+2\sqrt{7}\rangle$ has $\varphi(3)=0+\langle5+2\sqrt{7}\rangle$, meaning $3\in\ker\varphi$.
The problem is now to show that $\varphi$ is surjective. If you take $a+b\sqrt{7}\in\mathbb{Z}[\sqrt{7}]$ with even $b=2c$, then
$$
a+b\sqrt{7}=a+2c\sqrt{7}=(a-5c)+c(5+2\sqrt{7})
$$
and therefore $a+b\sqrt{7}+\langle5+2\sqrt{7}\rangle=\varphi(a-5c)$.
Can you work out the problem for odd $b$? Hint: compute $(5+2\sqrt{7})(-2+\sqrt{7})$.
Note. I used $\langle5+2\sqrt{7}\rangle$ to denote the ideal, in order to avoid confusions with parenthesized expressions in $\mathbb{Z}[\sqrt{7}]$.
