Using the AC Method to factor a trinomial

It has been ages since I've used the AC method and I can't quite remember how the negative numbers work.

The question is:
Solve the equation by factoring.
x^2-7x+12=0

When I look at the factors of 12, I get:
1,12
2,6
3,4

But none of those equal -7.
How do the negatives "work" when it comes to the factors above?

If the constant term (the 12 in this case) is positive, then you have to consider the case where both factors are positive (1,12 or 2,6 or 3,4) if the linear term (the -7 in this case) is also positive, OR the case where both factors are negative (-1,-12 or -2,-6, or -3,-4) if the linear term is negative. Because it's negative, both factors must be negative, so you need to consider (-1,-12 or -2,-6 or -3,-4). And when you do that, you find that -3 + -4 = -7. So the answer is:

(x - 4)(x- 3) = 0
x - 4 = 0 or x - 3 = 0
x = 4 or x = 3

3 and 4 don't equal -7, but -3 and -4 are also factors of +12 and do sum to -7

x²-7x+12 = (x-3)(x-4)