Solve 0<-16t^2+32t

$$0 < -16 { t }^{ 2 } +32t$$

Solve for t

$$t\in \left(0,2\right)$$

Solution Steps

Swap sides so that all variable terms are on the left hand side. This changes the sign direction.

$$-16t^{2}+32t>0$$

Multiply the inequality by -1 to make the coefficient of the highest power in -16t^{2}+32t positive. Since -1 is negative, the inequality direction is changed.

$$16t^{2}-32t<0$$

Factor out t.

$$16t\left(t-2\right)<0$$

For the product to be negative, t and t-2 have to be of the opposite signs. Consider the case when t is positive and t-2 is negative.

$$t>0 t-2<0$$

The solution satisfying both inequalities is t\in \left(0,2\right).

$$t\in \left(0,2\right)$$

Consider the case when t-2 is positive and t is negative.

$$t-2>0 t<0$$

This is false for any t.

$$t\in \emptyset$$

The final solution is the union of the obtained solutions.

$$t\in \left(0,2\right)$$