Describe $R=\mathbb{Z}[X]/(X^2-3,2X+4)$ 
I need to describe a ring $R=\mathbb{Z}[X]/(X^2-3,2X+4)$ 

I know that its element would be of the form $\{f(x)+(X^2-3,2X+4)|f(x)\in \mathbb{Z}[X]\}$
After that will i divide $f(x)$ by $X^2-3$ first and then by $2X+4$ to reduce the elements  in $R$? 
 A: it looks to me like it is $\mathbb Z_2[X]/\left<X^2+1\right>$.
$2$ is in your ideal because:
$$2 = 2(X^2-3)-(2X+4)(X-2)$$
Also $X^2+1$ is in your ideal because:
 $$\begin{align}X^2+1 &= (X^2-3) + 4 = (X^2-3) + 4(X^2-3) - (2X+4)(2X-4)  \\&= 5(X^2-3) - (2X+4)(2X-4) \end{align}$$
So we know that $\left<2,X^2+1\right>$ is contained in your ideal.
But it is obvious that the generators for your ideal are in $\left<2,X^2+1\right>$.
So we are done.
Note: Given that $(X+1)^2=X^2+1$ in $\mathbb Z_2$, this can be rewritten as isomorphic to $\mathbb Z_2[Y]/\left<Y^2\right>$ with $Y$ corresponding to $X+1$.
A: I have another answer.
$\mathbb Z[x]/(x^2-3,2x+4) = (\mathbb Z[x]/(x^2-3))/(x^2-3,2x+4)/(x^2-3)$(by third isomorphism theorem)
now define a map $\mathbb Z[x]$ to $\mathbb Z[\sqrt{3}], $
$F(f(x))=f(\sqrt{3})$( evaluated at root $3)$
then $F$ is surjective homomorphism.By 1st isomorphism theorem 
$\mathbb Z[x]/(x^2-3)=\mathbb Z[\sqrt{3}]$
and $(x^2-3,2x+4)/(x^2-3)=(2x+4+(x^2-3))$ and when it undergoes $F$,resultant ideal will be $(2\sqrt{3}+4)$
so finally description will be $\mathbb Z[\sqrt{3}]/(2\sqrt{3}+4)$
