well, in general if a quadratic can be factored, you can use the AC-method to factor it. do you know this method? we usually use it if the coefficient of x^2 is not 1, and is relatively big
here, 7 was not that big, so i decided to use trial and error. i know the only way to get 7 from multiplying integers is to have 7*1, so i set up the brackets
(7x )(x )
now, at the ends, i needed to put in numbers that i can multiply and get 12. 3 and 4 came to mind, so i tried them
(7x 3)(x 4)
now i needed to figure out the signs. we have -12, so the signs need to be different. moreover, i want the middle term to be -25x. but the middle term is calculated using the terms close together added to the terms far apart. that is, the middle term comes from some operation on 3x (=3*x) and 28x (=7x*4). if i have +3x and -28x then i get -25x, which is what i wanted, so i put those signs in accordingly.
(7x + 3)(x - 4)
of course, this is by trial and error, things won't always work out that smoothly, and you may have to switch numbers around and change numbers altogether. but since we are dealing with small numbers, it is not too much trouble
Read through this: Solving Quadratic Equations: Solving by Factoring