Quotient of a quotient ring

The ring is $$\mathbb{Z}[\sqrt{47}]/(2,1+\sqrt{47}) \cong(\mathbb{Z}[X]/(X^2-47))/(2,1+X) \cong\mathbb{Z}[X]/(X^2-47,2,1+X)\\ \cong \mathbb{Z}_2[X]/(X^2-47,1+X)= \mathbb{Z}_2[X]/(X^2-1,1+X) =\mathbb{Z}_2[X]/(1+X)\cong \mathbb{Z}_2$$ where $\mathbb{Z}_2 = \mathbb{Z}/(2)$.

How did you come to $(\Bbb Z[X]/(X^2 - 47))/(2, X^2 - 2X - 46)$ ?

It is $\cong \mathbb{Z}[\sqrt{47}]/(2,(1+\sqrt{47})^2)=\mathbb{Z}[\sqrt{47}]/(2) \cong \mathbb{Z}_2[X]/(1+X)^2$

Lets denote $R=\mathbb{Z}\left[\sqrt{47}\right]$, $I=\left<2\right>$ and $J=\left<1+\sqrt{47}\right>$. By the third isomorphism theorem

$$\mathbb{Z}\left[\sqrt{47}\right]/\left<2,1+\sqrt{47}\right>=R/\left(I+J\right)\cong\left(R/J\right)/\left(\left(I+J\right)/J\right)$$

As @Quasicoherent pointed out, I have made a mistake saying that $\mathbb{Z}\left[\sqrt{47}\right]/\left<1+\sqrt{47}\right>\cong\mathbb{Z}$. In fact

$$R/J=\mathbb{Z}\left[\sqrt{47}\right]/\left<1+\sqrt{47}\right>\cong\mathbb{Z}_{46}$$

since $\left<1+\sqrt{47}\right>\ni-\left(1-\sqrt{47}\right)\left(1+\sqrt{47}\right)=-\left(1-47\right)=46$ and

$$\left(a+b\sqrt{47}\right)\left(1+\sqrt{47}\right)=\left(a+47b\right)+\left(a+b\right)\sqrt{47}$$

is an integer iff $a=-b$ iff it is $46b$ for some $b$. Thus

$$\mathbb{Z}\left[\sqrt{47}\right]/\left<2,1+\sqrt{47}\right>\cong\mathbb{Z}_{46}/\left<2\right>=\left(\mathbb{Z}/46\mathbb{Z}\right)/\left(2\mathbb{Z}/46\mathbb{Z}\right)\cong\mathbb{Z}/2\mathbb{Z}=\mathbb{Z}_{2}$$

and $I+J=\left<2,1+\sqrt{47}\right>$ is prime.