Chapter 6 - Section 6.5 - Solving Equations Containing Rational Expressions - Exercise Set - Page 378: 48

$x=\dfrac{1}{10}$

Multiplying both sides by the $LCD= (2x-5)(2x+3) ,$ then the solution to the given equation, $ \dfrac{3}{2x-5}+\dfrac{2}{2x+3}=0 ,$ is \begin{array}{l}\require{cancel} (2x+3)(3)+(2x-5)(2)=0 \\\\ 6x+9+4x-10=0 \\\\ 10x-1=0 \\\\ 10x=1 \\\\ x=\dfrac{1}{10} .\end{array}