<s>[INST] $x^2 + 3x + 7 \equiv 0 \pmod {37}$ I'm trying to solve the following 
$x^2 + 3x + 7 \equiv 0 \pmod {37}$
What I've tried - 
I've tried making the left side as a square and then I know how to solve
but couldn't make it as a square root.. 
We also learned in class that you can multiply the left side and the modulo by $4a$ 
(that is $4\cdot 1 = 1$) and continue somehow - which I can't figure out how.
any help will be appreciated.
 [/INST] In the real numbers, a method of finding a solution to a quadratic equation is to complete the square. This would involve adding and subtracting $(b/2)^2$. $b=3$ in your case, and remember that $1/2 = 19 \mod 37$.
Specifically notice: $$(x+3 \cdot 19)^2 \equiv x^2 + 2\cdot  3 \cdot 19 x + (3 \cdot 19)^2$$ $$\equiv x^2 + 3x + (20)^2 \mod 37$$
Note that $3 \cdot 19 \equiv 20 \mod 37$. Also $20^2 = 400 \equiv 30 \mod 37$.
Thus the method of completing the square is as follows $$x^2 + 3x + 7 \equiv x^2 + 3x + 20^2 - 20^2 + 7 \equiv (x+20)^2 - 23 \mod 37$$
Finally this means you need to solve $$(x+20)^2 \equiv 23 \mod 37$$
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