Solve 5/2t^2-13t+12>0

$$\frac { 5 } { 2 } t ^ { 2 } - 13 t + 12 > 0$$

Solve for t

$$t\in \left(-\infty,\frac{6}{5}\right)\cup \left(4,\infty\right)$$

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.

$$\frac{5}{2}t^{2}-13t+12=0$$

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 \frac{5}{2} for a, -13 for b, and 12 for c in the quadratic formula.

$$t=\frac{-\left(-13\right)±\sqrt{\left(-13\right)^{2}-4\times \frac{5}{2}\times 12}}{2\times \frac{5}{2}}$$

Do the calculations.

$$t=\frac{13±7}{5}$$

Solve the equation t=\frac{13±7}{5} when ± is plus and when ± is minus.

$$t=4 t=\frac{6}{5}$$

Rewrite the inequality by using the obtained solutions.

$$\frac{5}{2}\left(t-4\right)\left(t-\frac{6}{5}\right)>0$$

For the product to be positive, t-4 and t-\frac{6}{5} have to be both negative or both positive. Consider the case when t-4 and t-\frac{6}{5} are both negative.

$$t-4<0 t-\frac{6}{5}<0$$

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

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

Consider the case when t-4 and t-\frac{6}{5} are both positive.

$$t-\frac{6}{5}>0 t-4>0$$

The solution satisfying both inequalities is t>4.

$$t>4$$

The final solution is the union of the obtained solutions.

$$t<\frac{6}{5}\text{; }t>4$$