Solve:
3n-7/(n-2)(n-3) + 2n+8/9-n^2 - n+2/(n+3)(n-2) = 0
Wow, I am sure you didn't mean what you typed.
Trusting my instincts, I think you meant
(3n-7)/[(n-2)(n-3)] + (2n+8)/(9-n^2) - (n+2)/[(n+3)(n-2)] = 0
(3n-7)/[(n-2)(n-3)] + 2(n+4)/[(3-n)(3+n)] - (n+2)/[(n+3)(n-2)] = 0
(3n-7)/[(n-2)(n-3)] - 2(n+4)/[(n-3)(3+n)] - (n+2)/[(n+3)(n-2)] = 0
the common denominator is
(n+3)(n-3)(n-2)
so [(3n-7)(n+3) - 2(n+4)(n-2) - (n+2)(n-3)]/[(n+3)(n-3)(n-2)] = 0
(3n-7)(n+3) - 2(n+4)(n-2) - (n+2)(n-3) = 0
Expand then simplify.
You will be amazed at how it breaks apart,
See if you can get n = 1.