<s>[INST] Quadratic congruence relation Is there a general formula for solving quadratic congruence relation like $ax^2-bx+c = 0$ in $\mathbb{Z}_{n}$ where $n \in \mathbb{N}$?
For example, I am trying to solve $x^2-3x+2 = 0$ in $\mathbb{Z}_{200}$. What kind of tools I need to solve it?
 [/INST] There are techniques, but no general formula. In this case there is a method, based on the factorization $(x-1)(x-2)$.
For any $x$, the numbers $x-1$ and $x-2$ are relatively prime. Note that $200=2^3\cdot 5^2$, and work modulo $8$ and $25$ separately.
Since $x-1$ and $x-2$ are relatively prime, we have $(x-1)(x-2)\equiv 0\pmod{8}$ if and only if $x\equiv 1\pmod{8}$ or $x\equiv 2\pmod{8}$.
Similarly,  we have $(x-1)(x-2)\equiv 0\pmod{25}$ if and only if $x\equiv 1\pmod{25}$ or $x\equiv 2\pmod{25}$.
That gives $4$ possibilities:
(i) $x\equiv 1\pmod{8}$, $x\equiv 1\pmod{25}$;
(i) $x\equiv 2\pmod{8}$, $x\equiv 2\pmod{25}$;
(iii) $x\equiv 1\pmod{8}$, $x\equiv 2\pmod{25}$;
(iv) $x\equiv 2\pmod{8}$, $x\equiv 1\pmod{25}$.
Using the Chinese Remainder Theorem, or inspection, we can then write down the $4$ solutions modulo $200$.
The solutions to (i) and (ii) are trivial. For (iii), the solution is (by inspection) $177$ (modulo $200$. We leave (iv) to you. 
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