number theory - $x^2 + 3x + 7 equiv 0 pmod {37}$

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.

Answer

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 $$(x+20)^2 \equiv 23 \mod 37$$