Identify $\mathbb{Z}[x]/(2x^2+1,2x-3)$. Identify $\mathbb{Z}[x]/(2x^2+1,2x-3)$.
I tried this: 
$$
\mathbb{Z}[x]/(2x^2+1,2x-3)=:R \\
(2x^2+1,2x-3)=:I \\
\because (2x^2+1)-x(2x-3)=3x+1, \\
2(3x+1)-3(2x-3)=11 \\
\therefore I=(2x^2+1,2x-3,11) \\
\therefore R \cong \mathbb{Z}_{11}[x]/(2x^2+1,2x-3) \\
\cong \mathbb{Z}_{11}[x]/(x^2+6,x-7) \\
\cong \mathbb{Z}_{11}
$$
but I'm totally unsure. Thank you for your help :)

$2x-3=0 \implies x=\frac 3 2$. Substituting $\frac {11} 2=0$, thus of course $11=0$. Our ring becomes $\mathbb Z_{11}[\frac 3 2]$. Now we have to check that we already have an element in $\mathbb Z_{11}$ such that multiplying it by $2$ it becomes $3$: $$2y \equiv 3 (mod \hspace{0,2cm}11)$$ we multply by $6$ and we get $$y \equiv 18 \equiv 7(mod \hspace{0,2cm}11)$$ and we already have $7 \in \mathbb Z_{11}$, thus $\mathbb Z_{11}[\frac 3 2] \simeq \mathbb Z_{11}$.
In the same way you could add the two equations and get $x^2 + x - 1=0$ and get $\mathbb Z_{11}[\frac{-1+\sqrt5}{2} ]$. The inverse of $2$ is already in $\mathbb Z_{11}$, it is $6$, and the square roots of $5$ are $(\pm 4)^2 \equiv 5$, so again $\mathbb Z_{11}[\frac{-1+\sqrt5}{2} ] \simeq \mathbb Z_{11}$.
