We note first that division by zero is not allowed and that for this reason \(4x+2\) may not be equal to zero and that therefore \(x=-{{1}\over{2}}\) is not a solution.

We now distinguish two cases, namely \(4x+2>0\) and \(4x+2<0\).
In both cases we multiply the inequality on both sides by \(4x+2\) because we then get a linear inequality, for which we know there is a solution method.

Suppose \(4x+2>0\), i.e. \(x> -{{1}\over{2}}\). Then we get \(6<-8(4x+2)\).
When we move everything with \(x\) to the left and all constant terms to the right, we get \(32x<-22\).
Then, dvision by the coefficient of \(x\)gives \(x < -{{11}\over{16}}\).
So we have the following system of inequalities: \(x> -{{1}\over{2}}\,\wedge\; x < -{{11}\over{16}}\)
and this simplifies to \(\text{an empty solution set}\).

Suppose \(4x+2<0\), i.e. \(x< -{{1}\over{2}}\). Then we get \(6>-8(4x+2)\).
When we move everything with \(x\) to the left and all constant terms to the right, we get \(32x>-22\).
Then, division by the coefficient of \(x\) gives \(x > -{{11}\over{16}}\).
So we have the following system of inequalities: \(x< -{{1}\over{2}}\,\wedge\; x > -{{11}\over{16}}\)
and this simplifies to \(-{{11}\over{16}}\lt x\lt -{{1}\over{2}}\).

The solution of the original inequality is \(-{{11}\over{16}}\lt x\lt -{{1}\over{2}}\).