Cardinality of the quotient ring $\mathbb{Z}[x]/(x^2-3,2x+4)$ This problem is from a practice exam I was working on.

What is the cardinality of the quotient $\mathbb{Z}[x]/(x^2-3,2x+4)$ ?

Thoughts. If I find a ring that is easier to handle then this then I can go from there. So I think this is isomorphic to $\mathbb{Z}_2[x]/(x^2-3,x+2)$. And then I am stuck.
Am I correct so far? What should I be looking to do from here?
Thanks for your time and your answers.
 A: Hint $\ $ Note $\,(2\!-\!x)(2\!+\!x) = (2\!-\!\sqrt 3)(2\!+\sqrt 3)=1\,$ in $\ \Bbb Z[x]/(\color{#0a0}{x^2\!-\!3})\cong \,\Bbb Z[\sqrt 3],\, $ therefore, since $\ 2\!+\!x\,$ in a unit $\!\bmod \color{#0a0}{x^2\!-\!3},\,$ we can cancel  it from the generator $\,\color{#c00}2(2\!+\!x)\, $ of $I,\,$ yielding
$\qquad\ \ I = (\color{#0a0}{x^2\!-\!3},\,\color{#c00}2(2\!+\!x)) = (x^2\!-\!3,\color{#c00}2) = ((x\!+\!1)^2,2),\  $ so $\,\ \Bbb Z[x]/I = \Bbb Z_2[x]/((x\!+\!1)^2)$

Remark $\ $ Alternatively we can do the cancellation purely equationally as below
$\bmod I\!:\,\ \color{#0a0}{x^2\equiv 3},\,\ \color{}2(\color{}{2\!+\!x)\equiv 0} \,\Rightarrow\,0\equiv \color{}2(2\!+\!x)(2\!-\!x) \equiv 2(4\!-\!\color{#0a0}{x^2})\equiv 2(4\!-\!\color{#0a0}3) \equiv \color{#c00}2\ $
More generally when studying algebraic number rings we can often deduce much  by taking norms (e.g. above we used: $\,\alpha = 2+\sqrt{3}\,$ is a unit since its norm $\,N(\alpha ) = \alpha \bar\alpha  = 1)$.

The resulting quotient ring is $\,\cong \Bbb Z_2[\epsilon]/(\epsilon^2),\,$ the ring of dual numbers over $\,\Bbb Z_2.\,$ Such dual number rings prove useful for algebraically modelling derivatives, tangent spaces, etc.

The hint amounts to Euclid's Lemma in ideal form, i.e.
Lemma $\,\ \color{#0a0}{(a,b)=1}\,\Rightarrow\, (a,bc) = (a,c)\ $
Proof $\ \ \ (a,bc) = (a,ac,bc) = (a,\color{#0a0}{(a,b)}c) = (a,c)$
A: Your start is good  as long as you can show that $2\in(x^2-3,2x+4)$. Fortunately this is the case: $x(2x+4)-2(x^2-3)=4x+6$ and $2(2x+4)-(4x+6)=2$. (Actually one can show that $(x^2-3,2x+4)=(x^2+1,2)$.) So $$\mathbb{Z}[x]/(x^2-3,2x+4)\simeq \mathbb{Z}_2[x]/(x^2+1)$$ and this ring has $4$ elements. 
A: A slight modification of YACP's answer:
If a ring contains an element $x$ satisfying $x^2=3$ and $2x=-4$, then we get $6=2x^2=-4x=8$, hence $2=0$, and the second relation becomes superflous. This shows that $\mathbb{Z}[x]/(x^2-3,2x+4)=\mathbb{Z}/2[x]/(x^2-3)$.
I just want to emphasize that you don't have to come up with clever linear combinations of the generators of the ideal. Just compute inside the quotient ring!
(In fact, as usual, you can also see this as an application of the Yoneda Lemma: The algebraic manipulations above show that $\hom(\mathbb{Z}[x]/(x^2-3,2x+4),-) \cong \hom(\mathbb{Z}/2[x]/(x^2-3),-)$.)
