Solve the inequality $|3x-5| - |2x+3| >0$. In order to solve the inequality $|3x-5| - |2x+3| >0$, I added $|2x+3|$ to both sides of the given inequality to get $$|3x-5| > |2x+3|$$ Then assuming that both $3x-5$ and $2x+3$ are positive for certain values of $x$, $$3x-5 > 2x+3$$ implies $$x>8$$ If $3x-5$ is positive and $2x-3$ is negative for certain values of $x$, then $$3x-5 > -2x-3$$ implies $$5x >2$$ implies $$x > \dfrac{2}{5}$$ I'm supposed to get that $x < \dfrac{2}{5}$ according to the solutions, but I'm not sure how to get that solution.

Needless to explicit the absolute values in function of the intervals where $x$ lives:
$$|3x-5| > |2x+3| \iff (3x-5)^2>(2x+3)^2\iff 5x^2-42x+16>0,$$
so it comes down to a quadratic inequality.
The reduced discriminant is $\Delta'=21^2-80=361=19^2$, and the quadratic  is positive outside the interval of the roots.
