Solve 2x^2-5x>0

$$2 x ^ { 2 } - 5 x > 0$$

Solve for x

$$x\in \left(-\infty,0\right)\cup \left(\frac{5}{2},\infty\right)$$

Solution Steps

Factor out x.

$$x\left(2x-5\right)>0$$

For the product to be positive, x and x-\frac{5}{2} have to be both negative or both positive. Consider the case when x and x-\frac{5}{2} are both negative.

$$x<0 x-\frac{5}{2}<0$$

The solution satisfying both inequalities is x<0.

$$x<0$$

Consider the case when x and x-\frac{5}{2} are both positive.

$$x-\frac{5}{2}>0 x>0$$

The solution satisfying both inequalities is x>\frac{5}{2}.

$$x>\frac{5}{2}$$

The final solution is the union of the obtained solutions.

$$x<0\text{; }x>\frac{5}{2}$$