Solve for x
x\in \left(-\infty,-\frac{3}{2}\right)\cup \left(2,\infty\right)
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2x+3<0 x-2<0
For the quotient to be positive, 2x+3 and x-2 have to be both negative or both positive. Consider the case when 2x+3 and x-2 are both negative.
x<-\frac{3}{2}
The solution satisfying both inequalities is x<-\frac{3}{2}.
x-2>0 2x+3>0
Consider the case when 2x+3 and x-2 are both positive.
x>2
The solution satisfying both inequalities is x>2.
x<-\frac{3}{2}\text{; }x>2
The final solution is the union of the obtained solutions.
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