Number of solutions in quadratic congruence

Question

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:

1. $x\equiv 2$ mod $7$ and $x\equiv 2$ mod $4$
2. $x\equiv 2$ 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?

Answer 1

Hint. Note that $$x^2+3x+18\equiv x^2+3x-10= (x+5)(x-2)\pmod{28}.$$

Answer 2

$x\equiv2\pmod4,x\equiv2\pmod7\implies$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 $(4,7)=1$

$\implies a+1=4c$

$\implies x=7a+2=7(4c-1)+2=28c-5\equiv-5\pmod{28}\equiv-5+28$