(2x - 5)/[(x-2) (3x-1) (2x+5)]
= A/(x-2) + B/(3x-1) + C/(2x+5) =>
2x - 5 = A(6x^2 + 13x - 5) + B(2x^2 + x - 10) + C(3x^2 - 7x + 2) =>
-5 = -5A - 10B + 2C;
2 = 13A + B - 7C;
0 = 6A + 2B + 3C.
Now you solve the three equations simultaneously the same way you would have done it in high school. I would start by eliminating "B" from the last two equations, obtaining 4 = 20A - 17C. Then by eliminating "B" from the first two equations, obtaining 15 = 125A - 68C. Then eliminate "C" from this pair, obtaining -1 = 45A. I can see my A = -1/45 in the given answer, but I can't guess where the 39 came from, hmm. Anyway, check my work and keep going from there.