How many elements does the ring $ℤ[X]/(X^2-3,2X-4)$ have? Describe the structure of this ring. I'm making exercises to prepare for my ring theory exam:

How many elements does the ring $ℤ[X]/(X^2-3,2X-4)$ have? Describe the
  structure of the this ring.

I find it always difficult if the ideal is generated by two elements. 
I thought about someting like this. $X=2$ and $X^2=3$. Therefore $4=3$ therefore $1=0$. As $2=1$ then $X=0$. And then everyting seems to become $0$. 
I'm not sure if I'm allowed to see it this way, but this is the first thing that comes to my mind. 
 A: Henry's answer does a good job of explaining what the congruence classes look like, except for a misstep about $b$. By using the Euclidean algorithm on $x^2-3$ and $2x-4$ in $\Bbb Q[x]$, we can find that $2(x^2-3)-(2x-4)(x+2)=2$, so $2$ is in that ideal, and hence $(x^2-3,2x-4)=(x^2+1,2)$. That further reduces the possibilities for $b$ to $0,1$ also. 
I'm going to add a bit about the structure of your ring (which I'll call $R$).
Clearly it is a commutative finite ring: $\{0,1,x,x+1\}$, and you can use an isomorphism theorem to show it is isomorphic to $\Bbb F_2[x]/(x+1)^2$
A: Here is another way to compute this.
In the quotient ring, we have $X^2 = 3$ (I willl continue to write $X$ to denote the image of $X$ in the quotient)).  Thus $(X-2)(X+2) = X^2 - 4 = - 1,$ and so in the quotient we have $X-2$ is a unit.  Thus the equation $2(X-2) = 0$ simplifies to $2 = 0$.   Thus the quotient is equal to $(\mathbb Z/2\mathbb Z)[X]/(X^2 -3)$.
Now $\mathbb Z/2\mathbb Z$ is a field of two elements, and in particular in this field $(a+b)^2 = a^2 + b^2$.  Using this, we see that $X^2 - 3 = X^2 + 1 = (X+1)^2$.  If we make a change of variables $T = X+ 1$, then we can write the quotient as $(\mathbb Z/2\mathbb Z)[T]/(T^2).$  
