Solve 3x^2+4x>0

$$3 x ^ { 2 } + 4 x > 0$$

Solve for x

$$x\in \left(-\infty,-\frac{4}{3}\right)\cup \left(0,\infty\right)$$

Solution Steps

Factor out x.

$$x\left(3x+4\right)>0$$

For the product to be positive, x+\frac{4}{3} and x have to be both negative or both positive. Consider the case when x+\frac{4}{3} and x are both negative.

$$x+\frac{4}{3}<0 x<0$$

The solution satisfying both inequalities is x<-\frac{4}{3}.

$$x<-\frac{4}{3}$$

Consider the case when x+\frac{4}{3} and x are both positive.

$$x>0 x+\frac{4}{3}>0$$

The solution satisfying both inequalities is x>0.

$$x>0$$

The final solution is the union of the obtained solutions.

$$x<-\frac{4}{3}\text{; }x>0$$