Solve $x^2+x+3=0$ mod $27$ I was preparing for my Number Theory class for next semester and one of the questions that I came upon is to solve $x^2+x+3=0$ mod $27$. I have seen modular arithmetic before but never one that involved using mod with an equation. I plugged it into WolframAlpha and it gave me $x=11,15$. My question is, how did wolfram alpha solve this? Is there some kind of procedure? My initial try was to use Brute Force and plug in every value possible until it is divisible by 27 but I would like to see how it is solved.

Your equation is equivalent to the following:
$$4x^2+4x+12\equiv0 \mod27$$ since $\gcd(4,27)=1$. Or $$(2x+1)^2+11\equiv0 \mod27$$ and even better $$(2x+1)^2-16\equiv0\mod27$$  which gives $27\mid (2x-3)(2x+5)$. But $\gcd(2x-3,2x+5)$ must be a divisor of $8$ which is relatively prime to $27$.  So, $27|2x-3$ or $27| 2x+5 $. The first case gives $x\equiv15$ and the second one gives $x\equiv11$ (mod27)
