solve this equation with fraction?

2x/(x + 3) - 5/(x + 1) = 0
[x + 3][x + 1][2x/(x + 3) - 5/(x + 1)] = 0
2x(x + 1) - 5(x + 3) = 0
2x^2 + 2x - 5x - 15 = 0
2x^2 - 3x - 15 = 0
x = [-b ±√(b^2 - 4ac)]/2a

a = 2
b = -3
c = -15

x = [3 ±√(9 + 120)]/4
x = [3 ±√129]/4

∴ x = [3 ±√129]/4

2x/x + 3 - 5/x + 1 = 0
2x/x - 5/x + 4 = 0
2x/x - 5/x = -4
(2x - 5)/x = -4
-4x = 2x - 5
4x = -(2x - 5)
4x = 5 - 2x
2x = 5
x = 5/2

2x^2 - 7x + 15
= 2(5/2)^2 - 7(5/2) + 15
= 2 (25/4) - 35/2 + 15
= 50/4 - 70/4 + 60/4
=180/4
=45

add the second term to the other side, then cross multiply
you get 2x^2+2x=5x+15
which means 2x^2-3x-15=0
This does not factor nicely. Use quadratic formula.
a=2, b=-3, c=-15

2x / (x + 3) - 5 / (x + 1) = 0

2x(x + 1) - 5(x + 3)
----------------------------
(x + 3)(x + 1)

2x² - 3x - 15
-----------------------
(x + 3)(x + 1)