We note first that division by zero is not allowed and that for this reason \(2x+8\) may not be equal to zero and that therefore \(x=-4\) is not a solution.

We now distinguish two cases, namely \(2x+8>0\) and \(2x+8<0\).
In both cases we multiply the inequality on both sides by \(2x+8\) because we then get a linear inequality, for which we know there is a solution method.

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

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

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