We note first that division by zero is not allowed and that for this reason \(5x+4\) may not be equal to zero and that therefore \(x=-{{4}\over{5}}\) is not a solution.

We now distinguish two cases, namely \(5x+4>0\) and \(5x+4<0\).
In both cases we multiply the inequality on both sides by \(5x+4\) because we then get a linear inequality, for which we know there is a solution method.

Suppose \(5x+4>0\), i.e. \(x> -{{4}\over{5}}\). Then we get \(2<-5(5x+4)\).
When we move everything with \(x\) to the left and all constant terms to the right, we get \(25x<-22\).
Then, dvision by the coefficient of \(x\)gives \(x < -{{22}\over{25}}\).
So we have the following system of inequalities: \(x> -{{4}\over{5}}\,\wedge\; x < -{{22}\over{25}}\)
and this simplifies to \(\text{an empty solution set}\).

Suppose \(5x+4<0\), i.e. \(x< -{{4}\over{5}}\). Then we get \(2>-5(5x+4)\).
When we move everything with \(x\) to the left and all constant terms to the right, we get \(25x>-22\).
Then, division by the coefficient of \(x\) gives \(x > -{{22}\over{25}}\).
So we have the following system of inequalities: \(x< -{{4}\over{5}}\,\wedge\; x > -{{22}\over{25}}\)
and this simplifies to \(-{{22}\over{25}}\lt x\lt -{{4}\over{5}}\).

The solution of the original inequality is \(-{{22}\over{25}}\lt x\lt -{{4}\over{5}}\).