 Solution

Write $f=X^2+bX+c$ and $g=X^2+BX+C$. Since their common square-free discriminant $d$ is congruent to zero or one modulo four, it must be one modulo four; thus $b,B$ are odd. It is not a rational square because a square-free integer square is only zero or one. Both polynomials are irreducible over $\mathbb{Q}$, and their roots generate the same quadratic extension $\mathbb{Q}(\sqrt{d})$. Their integral roots are $(-b+\sqrt{d})/2$ and $(-B+\sqrt{d})/2$, respectively. Each generated order isthe usual ring of integers for square-free $d\equiv 1(\mathrm{mod}4)$. This identifies both quotients with the same ring. Explicitly, if $\xi$ denotes the image of $X$ modulo $g$, send the image of $X$ modulo $f$ to $\xi +(B-b)/2$. Equal discriminants give $f(\xi +(B-b)/2)=g(\xi )=0$, and the opposite translation gives its inverse. The two quotient rings are isomorphic.