Solve {l}{25t^2}{-120t}{+108>0}

$$\left. \begin{array} { l } { 25 t ^ { 2 } } \\ { - 120 t } \\ { + 108 > 0 } \end{array} \right.$$

Solve for t

$$t\in \left(-\infty,\frac{6}{5}\right)\cup \left(\frac{18}{5},\infty\right)$$

Steps Using the Quadratic Formula

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.

$$25t^{2}-120t+108=0$$

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 25 for a, -120 for b, and 108 for c in the quadratic formula.

$$t=\frac{-\left(-120\right)±\sqrt{\left(-120\right)^{2}-4\times 25\times 108}}{2\times 25}$$

Do the calculations.

$$t=\frac{120±60}{50}$$

Solve the equation t=\frac{120±60}{50} when ± is plus and when ± is minus.

$$t=\frac{18}{5} t=\frac{6}{5}$$

Rewrite the inequality by using the obtained solutions.

$$25\left(t-\frac{18}{5}\right)\left(t-\frac{6}{5}\right)>0$$

For the product to be positive, t-\frac{18}{5} and t-\frac{6}{5} have to be both negative or both positive. Consider the case when t-\frac{18}{5} and t-\frac{6}{5} are both negative.

$$t-\frac{18}{5}<0 t-\frac{6}{5}<0$$

The solution satisfying both inequalities is t<\frac{6}{5}.

$$t<\frac{6}{5}$$

Consider the case when t-\frac{18}{5} and t-\frac{6}{5} are both positive.

$$t-\frac{6}{5}>0 t-\frac{18}{5}>0$$

The solution satisfying both inequalities is t>\frac{18}{5}.

$$t>\frac{18}{5}$$

The final solution is the union of the obtained solutions.

$$t<\frac{6}{5}\text{; }t>\frac{18}{5}$$