Determine $x$ such that $2x^2 - x - 6$ be a multiple of $53$ I used the modular arithmetic :
$$2x^2 - x - 6  \equiv 0  \mod 53$$
and then : 
$$(2x+3)(x-2) \equiv 0 \mod 53$$
I'm stuck here and don't know what to do with $AB \equiv 0  \mod C$
 A: Since $53$ is prime, it divides either $x-2$ or $2x+3$. Thus $x\equiv 2,\,25\,(\operatorname{mod}53)$.
A: It follows from your computations that $x=2$ will work.
If you're after all solutions (modulo $53$), you do\begin{align}(2x+3)(x-2)\equiv0\pmod{53}&\iff 2x+3\equiv0\pmod{53}\vee x-2\equiv0\pmod{53}\\&\iff2x\equiv50\pmod{53}\vee x\equiv2\pmod{53}\\&\iff x\equiv25\pmod{53}\vee x\equiv2\pmod{53}.\end{align}
A: See Euclid's Lemma. $53$ divides the product $(2x+3)(x-2)$. Then $53$ divides one of them, or both. You now have all the cases.
A: You can do also like this. Multiply this with $8$
$$2x^2 - x - 6  \equiv 0  \mod 53$$
so 
$$16x^2 - 8x - 48  \equiv 0  \mod 53$$
so
$$16x^2 - 8x +1- 49  \equiv 0  \mod 53$$
so
$$(4x-1)^2   \equiv 7^2  \mod 53$$
so 
$$4x-1   \equiv \pm 7  \mod 53$$
...

Note that if you have
$$ax^2+bx+c\equiv 0  \mod d$$
and $\gcd(2a,d) =1$ then you can do this always (multiply with $4a$):
$$4a^2x^2+4abx+4ac\equiv 0  \mod d$$
so
$$4a^2x^2+4abx+b^2\equiv b^2-4ac  \mod d$$
$$(2ax+b)^2\equiv \Delta   \mod d$$ where $\Delta = b^2-4ac$.
A: $53=1*53$(it is a prime number)    
So once let, $2x+3=53$
And second time let $x-2=53$
For (i) $x=25$
$\Rightarrow$ $Number=53*27$    
Similarly for (ii) $x=55$
$\Rightarrow$ $Number=53*113$    
Now considering another case when $53|2x+3$ or $53|x-2$
Which means (i) $53n=2x+3$ (Here n can take any odd integral value)
OR
(ii) $53n=x-2$(here n can take any integral value)  
It means there can't be any discrete value/s of x
