Solve {l}{2t^2+15t}{+7<0}

$$\left. \begin{array} { l } { 2 t ^ { 2 } + 15 t } \\ { + 7 < 0 } \end{array} \right.$$

Solve for t

$$t\in \left(-7,-\frac{1}{2}\right)$$

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.

$$2t^{2}+15t+7=0$$

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 2 for a, 15 for b, and 7 for c in the quadratic formula.

$$t=\frac{-15±\sqrt{15^{2}-4\times 2\times 7}}{2\times 2}$$

Do the calculations.

$$t=\frac{-15±13}{4}$$

Solve the equation t=\frac{-15±13}{4} when ± is plus and when ± is minus.

$$t=-\frac{1}{2} t=-7$$

Rewrite the inequality by using the obtained solutions.

$$2\left(t+\frac{1}{2}\right)\left(t+7\right)<0$$

For the product to be negative, t+\frac{1}{2} and t+7 have to be of the opposite signs. Consider the case when t+\frac{1}{2} is positive and t+7 is negative.

$$t+\frac{1}{2}>0 t+7<0$$

This is false for any t.

$$t\in \emptyset$$

Consider the case when t+7 is positive and t+\frac{1}{2} is negative.

$$t+7>0 t+\frac{1}{2}<0$$

The solution satisfying both inequalities is t\in \left(-7,-\frac{1}{2}\right).

$$t\in \left(-7,-\frac{1}{2}\right)$$

The final solution is the union of the obtained solutions.

$$t\in \left(-7,-\frac{1}{2}\right)$$