Solve 0<-16t^2+56t-24

$$0 < - 16 t ^ { 2 } + 56 t - 24$$

Solve for t

$$t\in \left(\frac{1}{2},3\right)$$

Steps Using the Quadratic Formula

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

$$-16t^{2}+56t-24>0$$

Multiply the inequality by -1 to make the coefficient of the highest power in -16t^{2}+56t-24 positive. Since -1 is negative, the inequality direction is changed.

$$16t^{2}-56t+24<0$$

To solve the inequality, factor the left hand side. Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.

$$16t^{2}-56t+24=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 16 for a, -56 for b, and 24 for c in the quadratic formula.

$$t=\frac{-\left(-56\right)±\sqrt{\left(-56\right)^{2}-4\times 16\times 24}}{2\times 16}$$

Do the calculations.

$$t=\frac{56±40}{32}$$

Solve the equation t=\frac{56±40}{32} when ± is plus and when ± is minus.

$$t=3 t=\frac{1}{2}$$

Rewrite the inequality by using the obtained solutions.

$$16\left(t-3\right)\left(t-\frac{1}{2}\right)<0$$

For the product to be negative, t-3 and t-\frac{1}{2} have to be of the opposite signs. Consider the case when t-3 is positive and t-\frac{1}{2} is negative.

$$t-3>0 t-\frac{1}{2}<0$$

This is false for any t.

$$t\in \emptyset$$

Consider the case when t-\frac{1}{2} is positive and t-3 is negative.

$$t-\frac{1}{2}>0 t-3<0$$

The solution satisfying both inequalities is t\in \left(\frac{1}{2},3\right).

$$t\in \left(\frac{1}{2},3\right)$$

The final solution is the union of the obtained solutions.

$$t\in \left(\frac{1}{2},3\right)$$