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