$ \frac{2x + 1}{x^2 - 8x + 16} + \frac{3x + 3}{2x^2 - 32} + \frac{12}{4-x} = 0 $

Problem:

$ \frac{2x + 1}{x^2 - 8x + 16} + \frac{3x + 3}{2x^2 - 32} + \frac{12}{4-x} = 0 $

Solution:

$ x = 5 $ or $ x = \frac{-76}{17} $