Solve $x^2+5x+6 \equiv 0 \pmod{\!11\cdot 17}$ Solve $x^2+5x+6 \equiv 187 \mod 187$
Solution
$$x^2+5x+6 \equiv 187 \mod 187$$
$$ (x+\frac{5}{2})^2 \equiv \frac{1}{4}$$
$$ 4(x+\frac{5}{2})^2 \equiv 1$$
$$ y:= x+\frac{5}{2} $$
$$ 4y^2 \equiv 1 \mod 11 \wedge 4y^2 \equiv 1 \mod 17  $$
$$ ( 2y \equiv 1 \mod 11 \vee 2y \equiv 10 \mod 11 ) \wedge ( 2y \equiv 1 \mod 17  \vee 2y \equiv 13 \mod 17)  $$
$$ ( y \equiv 6 \mod 11 \vee y \equiv 5 \mod 11 ) \wedge ( y \equiv 9 \mod 17  \vee y \equiv 15 \mod 17)  $$
Combining that from CRT I got:
$$ y \in \left\{49, 60,83,94 \right\} $$
and for example:
$$ x+\frac{5}{2}  \equiv 94 \mod 187$$
$$ 2x  \equiv 183 \mod 187$$
some calculus and get...
$$x \equiv 185 $$
And the same thing for each other case.
Question
Is there any faster (or smarter) way to solve equations like that?

Solve $x^2+5x+6 \equiv (x+2)(x+3) \pmod {187}$.
Besides the two 'in your face' solutions, $x \equiv -2 \pmod{187}$ and $x \equiv -3 \pmod{187}$, we can buttress the argument given by José Carlos Santos to find all four solutions.
We want to find integers $x, k, j$ satifying
$\;\text{L1:}\quad x + 2 = 11k$
$\;\text{L2:}\quad x + 3 = 17j$
Subtracting $\text{L1}$ from $\text{L2}$ we write
$\tag 1 1  = 17j - 11k$
Bézout's identity gives us
$\quad 1 = 17(2) - 11(3)$
So $k = 3$ and, plugging into $\text{L1}$,  $x \equiv 31 \pmod{187}$ is a solution.
For the last solution,
$\;\text{L1:}\quad x + 2 = 17j$
$\;\text{L2:}\quad x + 3 = 11k$
Subtracting $\text{L1}$ from $\text{L2}$ we write
$\tag 2 1  = -17j + 11k$
Bézout's identity gives us
$\quad 1 = 17(2) - 11(3)$
So $k = -3$ and, plugging into $\text{L2}$,  $x = -36 \equiv 151 \pmod{187}$  is a solution.
It can also be argued that there are exactly four solutions.
