solving modulus equation using triangle inequality

Determine the values of $x$ satisfying the equality $|3x^2-5|-|2x^2+3|=|x^2-8|$

I can see that it is written in the form $|x|-|y|=|x-y|$ which is true when $x$ and $y$ are of different signs so $xy\le0$ but that does not give the correct answer.

Please tell what am i doing wrong and what is the correct method to solve these kind of problems.

You are wrong about when the equality $|x|-|y|=|x-y|$ holds. It holds when (and only when) of of these conditions holds:
• $x\geqslant y\geqslant0$;
• $x\leqslant y\leqslant0$.
I shall now use this to solve your problem. The equality holds if and only if$$3x^2-5\geqslant2x^2+3\geqslant0\text{ or }3x^2-5\leqslant2x^2+3\leqslant0.$$The secons possibility cannot take place, of course. On the other hand\begin{align}3x^2-5\geqslant2x^2+3&\iff x^2\geqslant8\\&\iff x\geqslant\sqrt8\vee x\leqslant-\sqrt8.\end{align}
• so $x$ and $y$ should be of same sign, which implies $xy>0$ – Gem Aug 16 '17 at 14:52
• @Gem Yes, of course. Otherwise $|x-y|>\bigl||x|-|y|\bigr|\geqslant|x|-|y|$. – José Carlos Santos Aug 16 '17 at 14:55
• @Gem The solution is $x\geqslant\sqrt8\vee x\leqslant-\sqrt8$. Do you agree? – José Carlos Santos Aug 16 '17 at 15:03
• yes but how do you get it by solving $(3x^2−5)*(2x^2+3)>0$ – Gem Aug 16 '17 at 15:05