$(\mathbb{Z}/2\mathbb{Z})[x] /\langle x^2 -2\rangle \ncong (\mathbb{Z}/2\mathbb{Z})[x] /\langle x^2 -3\rangle$ today I have a problem.

Let $R_1=\mathbb{Z}_2[x] /\langle x^2 -2\rangle$ and $R_2=\mathbb{Z}_2[x] /\langle x^2 -3\rangle$

prove or disprove $R_1$ and $R_2$ are isomorphic.

I felt confuse because $x^2 =2$ and $x^2=3$ have solution in $\mathbb{Z}/2\mathbb{Z}$

I don't know what to do.
Both sets have four elements, namely the equivalence classes represented by all polynomials over $\mathbb{Z}_2$ of up to first order. Now make multiplication tables for both sets. If they are essentially the same (up to relabelling entries and reordering rows and columns), the rings are isomorphic. Otherwise they aren't.
I'm not good at editing.I'm sorry:(
First of all, 2 in Z2 is O, because 2 mod 2=0 
(the remainder of the division 2 / 2 is 0), then 3 in Z2 is 1 (3/2= 1 , remainder=1).
=>
            X^2 = X ^2 - 1
            -1 is 2-1=1 in Z2

You should know that -m in Zn is n-m.
=> X^2= X^2 + 1 | +1
   X^2 + 1=x^2+2 (2 is 0 in Z2)=>

=> X^2 +1= X^2 | + X ^2

X^2+X^2+1=X^2+ X^2

 X^2 (1+1) +1=X^2(1+1)

1+1 is 0 =&gt;

0+1=0


1=0 
False.
Your ecuation  doesn't have a solution. x ^2=2 means x^2=0, with the only solution x=0, and x^2=3 it means x^2=1, x=1. We have a contradiction: x=1=0 False