Looks good. My only comment is that if you're using mod 13, you should change the -4 to a 9.
I would like some help with this question:
Q: x^2+x+1=0 (mod 13)
I can't factorise the LHS as it stands, so I changed the 0 to 13, rearrange to give:
x^2+x-12=0 (mod 13)
(x+4)(x-3)=0 (mod 13)
x=-4 or 3 (mod 13)
Does my method work or have I gone wrong somewhere? Pointers would be greatly appreciated.
You can always manipulate your constant term to a place where there are solutions within the range, like you did. It's a matter of making your determinant, , nonnegative.
However, it is possible that you will end up with non-whole number solutions, which may or may not be okay, depending on how strict you are with your modular definitions.
Completing the square works fine too, but not everything has a square root.
For example, consider x^2 + 4x + 6 = 0. If we try to complete the square we get first x^2 + 4x = -6 or x^2 + 4x + 4 = -6 +4 or (x+2)^2 = -2, or equivalently (x+2)^2 = 11. But there no integer whose square is 11 mod 13, so this equation has no solution.