By CRT $7x^2+x\equiv -22\equiv 38 \bmod 60$ if and only if:
$x(7x+1)\equiv x(3x+1)\equiv 2 \bmod 4$
$x(7x+1)\equiv x(x+1) \equiv 2 \bmod 3$
$x(7x+1)\equiv x(2x+1) \equiv 3 \bmod 5$
Finding the solutions to each of these can be done by hand.
You want numbers such that:
$x\equiv 2$ or $3\bmod 4$
$x\equiv 1 \bmod 3$
$x\equiv 1 \bmod 5$.
By CRT there is one congruence when $x\equiv 2\bmod 4$ and another when $x\equiv 3\bmod 4$
I will solve the first of these, the other one is analogous. We use the algebraic approach explained in wikipedia:
$x\equiv2\bmod 4$
$x\equiv 1\bmod 3$
$x\equiv 1 \bmod 5$
We have $x=2+4t$ so $2+4t\equiv 1 \bmod 3$
so $4t\equiv 2 \bmod 3\implies t\equiv 2 \bmod 3\implies t=2+3k$. Therefore $x=2+4(2+3k)=10+12k$.
We now have $x=10+12k\equiv 1 \bmod 5\implies 12k\equiv 1 \bmod 5\implies 2k\equiv 1\bmod 5\implies k\equiv 3 \bmod 5$. therefore $k=5s+3$
So $x=10+12(5s+3)=60s+46$. So $x\equiv 46 \bmod 60$
You were correct.
$$x^2\equiv 24\pmod{\! 60}\iff \begin{cases}x^2\equiv 24\equiv 0\pmod{\! 3}\\ x^2\equiv 24\equiv 0\pmod{\! 4}\\ x^2\equiv 24\equiv 4\pmod{\! 5}\end{cases}$$
$$\iff \begin{cases}x\equiv 0\pmod{\! 3}\\ x\equiv 0\pmod{\! 2}\\ x\equiv \pm 2\pmod{\! 5}\end{cases}$$
If and only if at least one of the two cases holds:
$1)$ $\ x\equiv 0\pmod{\! 6},\ x\equiv 2\pmod{\! 5}$
$2)$ $\ x\equiv 0\pmod{\! 6},\ x\equiv -2\pmod{\! 5}$
You can use Chinese Remainder theorem as follows (when I create new variables, they're integers):
$$x\equiv 0\pmod{\! 6}\iff x=6k$$
$$1)\ \ \ x\equiv 2\pmod{\! 5}\iff \color{#00F}6k\equiv \color{#00F}1k\equiv 2\pmod{\! 5}$$
$x=6(5n+2)=30n+12$.
$$2)\ \ \ x\equiv 3\pmod{\! 5}\iff \color{#00F}6k\equiv \color{#00F}1k\equiv 3\pmod{\! 5}$$
$x=6(5n+3)=30n+18$.
Another way you can use CRT (which is basically just finding an $x$ that works in $[0,30)$):
$1)\ \ \ (x\equiv 0\equiv 12\pmod{\! 6}$ and $x\equiv 2\equiv 12\pmod{\! 5})\iff x\equiv 12\pmod{\! 6\cdot 5},$
because (since $(6,5)=1$):
$$6,5\mid x-12\iff 6\cdot 5\mid x-12$$
Using this, in case $2)$ in the same way you find that $18$ works ($18\equiv 0\pmod{\! 6}$ and $18\equiv 3\pmod{\! 5}$).
So you have the congruence holds iff $x=30m\pm 12$ for some $m\in\Bbb Z$.
Best Answer
You use the quadratic formula!
No, really. But you need to interpret the terms correctly: rather than "dividing" by $2a$ (here, $2$) you need to multiply by a number $r$ such that $2r\equiv 1\pmod{11}$ (namely, $r=6$). And rather than trying to find a square root, here $\sqrt{b^2-4ac} = \sqrt{1-4} = \sqrt{-3}$, you want to find integers $y$ such that $y^2\equiv -3\pmod{11}$. It may be impossible to do so, but if you can find them, then plugging them into the quadratic formula will give you a solution; and if you cannot find them, then there are no solutions.
Now, as it happens, $-3$ is not a square modulo $11$; so there are no solutions to $x^2+x+1\equiv 0\pmod{11}$. (You can find out if $-3$ is a square by using quadratic reciprocity: we have that $-1$ is not a square modulo $11$, since $11\equiv 3\pmod{4}$. And since both $3$ and $11$ are congruent to $3$ modulo $4$, we have $$\left(\frac{-3}{11}\right) = \left(\frac{-1}{11}\right)\left(\frac{3}{11}\right) = -\left(-\left(\frac{11}{3}\right)\right) = \left(\frac{11}{3}\right) = \left(\frac{2}{3}\right) = -1,$$ so $-3$ is not a square modulo $11$).
(You can also verify that this is the case by plugging in $x=1,2,\ldots,10$ and seeing that none of them satisfy the equation).
On the other hand, if your polynomial were, say $x^2+x-1$, then the quadratic formula would say that the roots are $$\frac{-1+\sqrt{1+4}}{2}\qquad\text{and}\qquad \frac{-1-\sqrt{1+4}}{2}.$$ Now, $5$ is a square modulo $11$: $4^2 = 16\equiv 5\pmod{11}$. So we can take $4$ as one of the square roots, and taking "multiplication by $6$" as being the same as "dividing by $2$" (since $2\times 6\equiv 1\pmod{11}$), we would get that the two roots are $$\begin{align*} \frac{-1+\sqrt{5}}{2} &= \left(-1+4\right)(6) = 18\equiv 7\pmod{11}\\ \frac{-1-\sqrt{5}}{2} &= \left(-1-4\right)(6) = -30\equiv 3\pmod{11} \end{align*}$$ and indeed, $(7)^2 + 7 - 1 = 55\equiv 0\pmod{11}$ and $3^2+3-1 = 11\equiv 0\pmod{11}$.
We can definitely use this method when $2a$ is relatively prime to the modulus; if the modulus is not a prime, though, nor an odd prime power, then there may be more than $2$ square roots for any given number (or none). But for odd prime moduli, it works like a charm.