Modular Arithmetic - Quadratic Solutions Problem I've just been given the following question in my crypto class, and I think I'm fairly sorted for it, but I was just wondering whether there might be any extra solutions to the ones I've worked out.

Compute all solutions of $x^2 + 4x - 21 \equiv 0\,\bmod\,33$

First, I factorised this equation to give $(x + 7)(x - 3)$, which gives me the solutions $-7$ and $3$. However, under the conditions of modular arithmetic, I know that adding or subtracting $33$ as many times as we like will also provide an answer of zero. 
IE: Let's try x = 26. $$(26 + 7)(26 - 3) = 33 \times 23 \equiv 0\,\bmod\, 33$$
Thus, it becomes fairly obvious to see that solutions to this equation will take the form $[3]$ and $[26]$, where the square brackets denote congruence classes.
I was given the hint in class that we should try to make the brackets equal to the factors of 33 - IE; try and get to $11 \times 3$, for example. But I really can't see how this would work.
Any further input would be great, thank you!!

$(x+7)(x-3)\equiv0\mod 33$
Now, as you noticed $ x=-7,3$ are obvious solutions. Also, any number congrunet to $-7$ or $3$ modulo $33$ will be a solution, for instance, $26$, as it is congruent to $-7$ modulo $33$, or $36$, congruent to $3$ modulo $33$. 
But there are other possible solutions, congruent to neither $-7$ nor $3$ modulo $33$
If we can solve $x+7=11k$ and $x-3=3k$, for integers $x,k$ then we have a solution, since one of the factors is divisible to $11$ and the other divisible to $3$. 
Similarly, we can solve $x+7=3k$ and $x-3=11k$,under the same constraints we have another solution.
Incidentally, both these sets of equations have solutions in integers, namely $15$ and $14$.
Thus all solutions of $x$ must be congruent to one of $-7,3,15,14$ modulo $33$
