Number of roots of a quadratic polynomial with coefficients in ring $\mathbb{Z_{18}}$ I am trying to solve the following problem:   

How many roots can a polynomial $P(x) = ax^2 + bx + c $, where $a$, $b$, $c \in \mathbb{Z_{18}}$, have?
  ($\mathbb{Z_{18}} = \{0, 1, 2, \ldots, 17\}$)

Obviously, they can not have more than 18 roots. 
If it was in real numbers the answer would be either 2 or 0 as real polynomials have as many roots (can be pairs of complex numbers) as is their exponent.
I tried this...
Let $x_1$ be a root of $p(x)$.
Another root $x_2$ may exist;
$ 0 = a(x_1)^2 + bx_1 + c = a(x_2)^2 + bx_2 + c$
$ a(x_1)^2 + bx_1 = a(x_2)^2 + bx_2$
$ a((x_1)^2 - (x_2)^2) = b(x_2 - x_1)$
I can't just say $(x_1)^2 - (x_2)^2 = 0$ as there are zero divisors in $\mathbb{Z_{18}}$.
Is any of this in the right direction? I could use some help.

And is there a more general way I could describe possible roots of polynomials from that ring? 

 A: [I assume that $a$ is required to be nonzero so that we really have a quadratic; taking the problem statement literally you could just let $a=b=c=0$ to trivially get $18$ roots.]
The way I would approach this is using the Chinese remainder theorem.  Since $\mathbb{Z}_{18}\cong \mathbb{Z}_2\times\mathbb{Z}_9$, we can think about the problem separately in $\mathbb{Z}_2$ and $\mathbb{Z}_9$.
Now $\mathbb{Z}_2$ is easy, since it only has two elements: $x(x-1)=x^2-x$ has two roots, and that is the maximum possible.
Notice now that any quadratic $ax^2+bx+c$ over $\mathbb{Z}_{18}$ which reduces to $x^2-x$ mod $2$ will have $a\neq 0$.  This means that we are free to have $a=b=c=0$ when we reduce mod $9$, so that over $\mathbb{Z}_9$ we will have $9$ roots.  That will get us a quadratic over $\mathbb{Z}_{18}$ with all $18$ elements as roots!
Explicitly, we want $ax^2+bx+c$ which reduces to $x^2-x$ mod $2$ and reduces to $0$ mod $9$.  We can do this with $a=b=9$ and $c=0$, so the quadratic $9x^2+9x$ has $18$ roots over $\mathbb{Z}_{18}$.
