<s>[INST] Number of solutions in quadratic congruence I use an example to explain my question:

How many solutions are there to $x^2+3x+18\equiv 0$ (mod $28$).

Usually I will encounter these kind of problems. My first step must transform this equation to become 2 equations:
$$x^2+3x+18\equiv 0 \quad (\text{mod }7) $$
$$x^2+3x+18\equiv 0\quad (\text{mod }4) $$
Then, I just plug in the numbers $0$ to $6$ to eq 1 to find out there is one solution $x\equiv 2$ mod $7$.
For equation 2, I do the same thing, plug in $0$ to $3$ and there are two solutions:$x\equiv 2$ and $x\equiv 3$ mod $4$.
Then how should I continue?? Using Chinese Remainder Theorem? Because I have made the original equation into a two simultaneous equations, maybe something like:


*

*$x\equiv 2$ mod $7$ and $x\equiv 2$ mod $4$

*$x\equiv 2$ mod $7$ and $x\equiv 3$ mod $4$
Is it the general way to do it? But is there a faster way to determine the number of solutions without performing CRT?

 [/INST] $x\equiv2\pmod4,x\equiv2\pmod7\implies$lcm$(4,7)|(x-2)\implies x\equiv2\pmod{28}$
For the second, $$7a+2=4b+3\iff7a=4b+8-7\iff\dfrac{7(a+1)}4=b+2$$ which is an integer
$\implies4|7(a+1)\iff4|(a+1)$ as $(4,7)=1$
$\implies a+1=4c$
$\implies x=7a+2=7(4c-1)+2=28c-5\equiv-5\pmod{28}\equiv-5+28$
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