Modular equation with $x^2$

For $x \in \mathbb{Z}_{200}$solve this modular equation $$(x-1)(x-2) \equiv 0 \mod 200$$ I don't know how to deal with that $x$ occurs in second power, I mean $x^2$

Hint: $200 = 2^3 \times 5^2$. Solve first mod $2$ and mod $5$, then lift to $2^3$ and $5^2$, then Chinese remainder...
I would recommend that you do not expand your parentheses, that is, don't write it as $x^2 - 3x + 2 \equiv 0$. Rather, look at what your problem is telling you: You have two consecutive integers, and their product is divisible by $200 = 2^3 \cdot 5^2$. That means that of the two numbers, one must be divisible by $25$, and one must be divisible by $8$. It's possible that one of the numbers is both, in which case your product is either $1\cdot 0$ or $0\cdot 199$, i.e. $x = 2$ or $x = 1$.
The other case is that one of them is divisible by $25$ and the other is divisible by $8$. There are a two possibilities: $x = 26$ gives $25\cdot 24$, and $x = 177$ gives $176\cdot 175$. There are no other multiples of $25$ (below $200$) which are one away from a multiple of $8$.