For example $x^2 \equiv 13 \bmod 29$. The only solution that comes to mind is to just calculate the squares of all of the numbers from 0 to 28, but that's tedious. Is there a better method?
EDIT: the question that this is marked as a duplicate to doesn't contain an answer as to how to solve such congruences, only for determining whether there exists a solution. So while a similar question has been asked, it hasn't been properly answered and IMO it's not duplicate then.
One technique I've found for this is to try to complete the difference of squares. For example, consider $x^2\equiv24\bmod{125}$. We can notice that $24\equiv1024 \bmod 125$, so we have $x^2-1024\equiv(x-32)(x+32)\equiv0\bmod125$ and from that it easily follows that $x\equiv32\bmod 125$ or $x\equiv-32\equiv93\bmod 125$. We can eliminate the possibility of $(x-32)(x+32)$ being a number that is divisible by 125 because the two numbers differ by 64, and they would both have to be divisible by 5 for their product to be divisible by 125.