How does the quotient ring $\Bbb Z[x]/(x^2-x,4x+2)$ look like? 
How does the quotient ring $\Bbb Z[x]/(x^2-x,4x+2)$ look like?

Normally to solve this you play around with generators until you get something you can work with. I was unsuccessful in reducing it to something neat.
I have proven that $6$ is the smallest integer in the ideal. Thus we get $\Bbb Z[x]/(6,x^2-x,4x+2)$ I dont see much further simplification. I would want to get rid of $x^2$ but i do not see  how.
 A: We have $8(x^2-x)-2x(4x+2)=-12x=-3(4x+2)+6$, so $6\in (x^2-x,4x+2)$ ,and $(x^2-x,4x+2)=(6,x^2-x,4x+2)$.
Hence, your ring $R$ is isomorphic to $\mathbb{Z}/6 [x]/(x^2-x,-\bar{2}x+\bar{2})$.
The map given by the Chinese remainder theorem yields an isomorphism $R\simeq \mathbb{F}_2[x]/(x^2-x)\times\mathbb{F}_3[x]/(x^2-x,x-\bar{1})=\mathbb{F}_2[x]/(x^2-x)\times\mathbb{F}_3[x]/(x-\bar{1})$, which finally yields $$R\simeq\mathbb{F}_2\times\mathbb{F}_2\times\mathbb{F}_3.$$
If we follow carefully the proof, we get an explicit isomorphism.
Let $f$ be the ring morphism $f:R\to P\in\mathbb{Z}[x]\mapsto ([P(0)]_2,[P(1)]_2, [P(1)]_3)\in\mathbb{F}_2\times\mathbb{F}_2\times \mathbb{F}_3$.
It is easy to see $(x^2-x,4x+2)$ lies in the kernel of $f$. The induced map is then the desired isomorphism.
We can also prove this last fact directly, just to double check that everything is correct.
Claim. $f$ is surjective.
Given $a,b,c\in\mathbb{Z}$, we need to find $P\in\mathbb{Z}[x]$ such that $P(0)\equiv a \ [2], P(1)\equiv b \ [2]$ and $P(1) \equiv c \ [3]$.
We can try $P=ux+v$. We can choose $v=a$, $u=(b-a)+2k$, so the two first equations are satisfied. Now we want $(b-a)+2k+a\equiv c \ [3]$, and we take $k=b-c$. To sum up $P=(3b-a-2c)x+a$ does the job.
Claim. $\ker(f)=(x^2-x,4x-2)$.
As we said before, one inclusion is clear, so let $P\in\mathbb{Z}[x]$ such that $f(P)$ is trivial.  We want to prove that $P\in (x^2-x,4x+2)$. Dividing by $x^2-x$, and replacing $P$ by the corresponding remainder,one may assume that $P=ux+v$.
By assumption, $v$ is even, and $u+v$ is a multiple of $2$ and $3$, so $v=2m$ and $u+v=6n$, that is $u=6n-v=6n-2m$. Hence $P=-2mx+ 6nx+2m=m(2-2x)-n 6x$.
Now $6x$ lies in our ideal (since $6$ does), and $2-2x=4x+2-6x$ also lies in our ideal, so we are done.
Now apply the first isomorphism theorem.
A: We have
$$(x^2-x,4x+2)=(x,4x+2)(x-1,4x+2)=(2,x)(6,x-1).$$
The ideals $(2,x)$ and $(6,x-1)$ are comaximal, and by CRT $$\mathbb Z[x]/(x^2-x,4x+2)\simeq\mathbb Z[x]/(2,x)\times\mathbb Z[x]/(6,x-1)\simeq\mathbb Z/2\mathbb Z\times\mathbb Z/6\mathbb Z.$$
