$(1)$ Find all $x$, that solve $7x^2 + x + 22 \equiv 0 \pmod{60}$.
I tried to solve this by first considering the prime factorization $60 = 2^2\cdot 3\cdot 5$ and then using the Chinese Remainder Theorem, that is for $n_1, ..., n_k$ coprime and $a_1, ..., a_k$ there exists a $x \in \mathbb{Z}$ that solves the system of simultaneous congruences
\begin{cases} x \equiv a_1 & \pmod{n_1} \\ \quad \cdots \\ x \equiv a_k &\pmod{n_k} \end{cases}
In this case i just tried to find a solution to $(1)$ mod $3$, mod $4$, mod $5$ one at a time by testing $0,1,2$, $0,1,2,3$ and $0,1,2,3,4$ respectively. While all of the congruences have solutions I couldn't find a single $x \in \mathbb{Z}$ that solves all of them. However, Wolfram-Alpha says there are two solutions $x_1 = 31$ and $x_2 = 46$. So how do I apply the CRT to this in the correct way?