Question

does anyone know how to do this?

Answer 1

You're looking for values to factor this equation into (x+p)(x+q)
such that pq=12 and p+q=b

Assuming we're looking for simple values, you can start by factorizing 12:

12 = 3*4 = 2*6 = 12*1 = -3*-4 = -2*-6 = -12*-1

For 3 and 4, b would be 7
2 and 6 => b=8
12 and 1 => b=13

and for the negatives

b can be -7, -8 or -13

So the total set is -13, -8, -7, 7, 8, 13

Answer 2

It is factorable if you can solve it when it equals 0. 
There are many ways to do this. 

X^2+bx+12=0 

$(-b +- \sqrt{ b^{2} -4*12 } )/2$

This equation gives a good answer if 
$b ^{2} \geq 48$