Solve -t^2+30t-105>0

$$- t ^ { 2 } + 30 t - 105 > 0$$

Solve for t

$$t\in \left(15-2\sqrt{30},2\sqrt{30}+15\right)$$

Steps Using the Quadratic Formula

Multiply the inequality by -1 to make the coefficient of the highest power in -t^{2}+30t-105 positive. Since -1 is negative, the inequality direction is changed.

$$t^{2}-30t+105<0$$

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.

$$t^{2}-30t+105=0$$

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 1 for a, -30 for b, and 105 for c in the quadratic formula.

$$t=\frac{-\left(-30\right)±\sqrt{\left(-30\right)^{2}-4\times 1\times 105}}{2}$$

Do the calculations.

$$t=\frac{30±4\sqrt{30}}{2}$$

Solve the equation t=\frac{30±4\sqrt{30}}{2} when ± is plus and when ± is minus.

$$t=2\sqrt{30}+15 t=15-2\sqrt{30}$$

Rewrite the inequality by using the obtained solutions.

$$\left(t-\left(2\sqrt{30}+15\right)\right)\left(t-\left(15-2\sqrt{30}\right)\right)<0$$

For the product to be negative, t-\left(2\sqrt{30}+15\right) and t-\left(15-2\sqrt{30}\right) have to be of the opposite signs. Consider the case when t-\left(2\sqrt{30}+15\right) is positive and t-\left(15-2\sqrt{30}\right) is negative.

$$t-\left(2\sqrt{30}+15\right)>0 t-\left(15-2\sqrt{30}\right)<0$$

This is false for any t.

$$t\in \emptyset$$

Consider the case when t-\left(15-2\sqrt{30}\right) is positive and t-\left(2\sqrt{30}+15\right) is negative.

$$t-\left(15-2\sqrt{30}\right)>0 t-\left(2\sqrt{30}+15\right)<0$$

The solution satisfying both inequalities is t\in \left(15-2\sqrt{30},2\sqrt{30}+15\right).

$$t\in \left(15-2\sqrt{30},2\sqrt{30}+15\right)$$

The final solution is the union of the obtained solutions.

$$t\in \left(15-2\sqrt{30},2\sqrt{30}+15\right)$$