Solve -18x^2+18x<0

$$-18 { x }^{ 2 } +18x < 0$$

Solve for x

$$x\in \left(-\infty,0\right)\cup \left(1,\infty\right)$$

Solution Steps

Multiply the inequality by -1 to make the coefficient of the highest power in -18x^{2}+18x positive. Since -1 is negative, the inequality direction is changed.

$$18x^{2}-18x>0$$

Factor out x.

$$18x\left(x-1\right)>0$$

For the product to be positive, x and x-1 have to be both negative or both positive. Consider the case when x and x-1 are both negative.

$$x<0 x-1<0$$

The solution satisfying both inequalities is x<0.

$$x<0$$

Consider the case when x and x-1 are both positive.

$$x-1>0 x>0$$

The solution satisfying both inequalities is x>1.

$$x>1$$

The final solution is the union of the obtained solutions.

$$x<0\text{; }x>1$$