What is the ring $\Bbb Z[x]/(x^2-3,2x+4)$? How to describe an element of the Ring $\Bbb Z[x]/(x^2-3,2x+4)$? 
Or is it isomorphic to any well known ring?  
What is meant by quotienting $\Bbb Z[x]$ by a kernel like that?
 A: Hint $\rm\ \ \ mod\,\ (\color{#0a0}{3\!-\!x^2},\ \color{#c00}{2x\!+\!4})\!: $ 
$\quad 6\!\!\!\!\! \color{#0a0}{\ \overset{ \overset{\ \ \ \ \Large{3\ \ \equiv\ \ x^2}}{\phantom{M}} }\equiv}\!  $ $ 2x^2\!\!\!\color{#c00}{\ \overset{ \overset{\ \Large{2x\ \ \equiv\ \ -4}}{\phantom{M}} }\equiv}\!\! $ $ -4x \!\!\!\!\! \color{#c00}{\ \overset{ \overset{\ \Large{-2x\ \ \equiv\ \ 4\ \ \ \ \ \ }}{\phantom{M}} }\equiv}\!\!\!\!\!\! 8 \,\ \Rightarrow\,\ 2\equiv 0$
So $\rm\ \ 2 \in I = (x^2\!-\!3,\, 2x\!+\!4)\ \Rightarrow\ I = (2,I) = (2,\, I\ mod\ 2) = (2, x^2\!+\!1) = (2,(x\!+\!1)^2)$
So $\rm\ \Bbb Z[x]/I \,\cong\, \Bbb Z[x]/(2,(x\!+\!1)^2)\,\cong\,\Bbb Z_2[x]/((x+1)^2)\,\cong\, \Bbb Z_2[x]/(x^2)\,\cong\: $  dual numbers over $\,\Bbb Z_2$
Remark $\ $ The elimination process used to deduce $\,2\equiv 0\,$ is not ad-hoc. Rather it is a special case of a normal form algorithm for such ideals, e.g. see the discussion on Hermite and Smith normal forms in Henri Cohen's A Course in Computational Number Theory. 
A: It is the universal ring containing an element $x$ such that $x^2=3$ and $2x=-4$. It follows
$6=2x^2=-4x$
and on the oher hand $4x=-8$, hence $2=0$. But then the relations simplify to $0=x^2-1=(x+1)^2$. Thus, we get
$\mathbb{F}_2[x]/(x+1)^2$
This cannot be simplified anymore (except for the trivial substitution $y:=x+1$). It is the unique $\mathbb{F}_2$-algebra with $4$ elements with a nontrivial nilpotent element (the other ones being $\mathbb{F}_2 \times \mathbb{F}_2$ and $\mathbb{F}_4=\mathbb{F}_2[x]/(x^2+x+1)$).
Edit: The whole point of my answer is that you don't have to calculate with ideals and residue classes ... just compute in the quotient ring, then everything is simple.
