Example 1 Find the solution to the congruence equation $4 x^{2}+27 x-12 \equiv 0(\bmod 15)$.
Solve the absolute minimal complete residue system modulo 15: $-7,-6, \cdots,-1,0,1$, $2, \cdots, 7$. Direct calculation shows that $x=-6,3$ are solutions. Therefore, the solutions to this congruence equation are
$$x \equiv-6,3(\bmod 15),$$

The number of solutions is 2.