Solve 0.25x^2+0.5x>0

$$0.25 { x }^{ 2 } +0.5x > 0$$

Solve for x

$$x\in \left(-\infty,-2\right)\cup \left(0,\infty\right)$$

Solution Steps

Factor out x.

$$x\left(0.25x+0.5\right)>0$$

For the product to be positive, x+2 and x have to be both negative or both positive. Consider the case when x+2 and x are both negative.

$$x+2<0 x<0$$

The solution satisfying both inequalities is x<-2.

$$x<-2$$

Consider the case when x+2 and x are both positive.

$$x>0 x+2>0$$

The solution satisfying both inequalities is x>0.

$$x>0$$

The final solution is the union of the obtained solutions.

$$x<-2\text{; }x>0$$