We note first that division by zero is not allowed and that for this reason \(8x+6\) may not be equal to zero and that therefore \(x=-{{3}\over{4}}\) is not a solution.

We now distinguish two cases, namely \(8x+6>0\) and \(8x+6<0\).
In both cases we multiply the inequality on both sides by \(8x+6\) because we then get a linear inequality, for which we know there is a solution method.

Suppose \(8x+6>0\), i.e. \(x> -{{3}\over{4}}\). Then we get \(2<-6(8x+6)\).
When we move everything with \(x\) to the left and all constant terms to the right, we get \(48x<-38\).
Then, dvision by the coefficient of \(x\)gives \(x < -{{19}\over{24}}\).
So we have the following system of inequalities: \(x> -{{3}\over{4}}\,\wedge\; x < -{{19}\over{24}}\)
and this simplifies to \(\text{an empty solution set}\).

Suppose \(8x+6<0\), i.e. \(x< -{{3}\over{4}}\). Then we get \(2>-6(8x+6)\).
When we move everything with \(x\) to the left and all constant terms to the right, we get \(48x>-38\).
Then, division by the coefficient of \(x\) gives \(x > -{{19}\over{24}}\).
So we have the following system of inequalities: \(x< -{{3}\over{4}}\,\wedge\; x > -{{19}\over{24}}\)
and this simplifies to \(-{{19}\over{24}}\lt x\lt -{{3}\over{4}}\).

The solution of the original inequality is \(-{{19}\over{24}}\lt x\lt -{{3}\over{4}}\).