Solve t^2-33t-70leq0

$$t ^ { 2 } - 33 t - 70 \leq 0$$

Solve for t

$$t\in \begin{bmatrix}-2,35\end{bmatrix}$$

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.

$$t^{2}-33t-70=0$$

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, -33 for b, and -70 for c in the quadratic formula.

$$t=\frac{-\left(-33\right)±\sqrt{\left(-33\right)^{2}-4\times 1\left(-70\right)}}{2}$$

Do the calculations.

$$t=\frac{33±37}{2}$$

Solve the equation t=\frac{33±37}{2} when ± is plus and when ± is minus.

$$t=35 t=-2$$

Rewrite the inequality by using the obtained solutions.

$$\left(t-35\right)\left(t+2\right)\leq 0$$

For the product to be ≤0, one of the values t-35 and t+2 has to be ≥0 and the other has to be ≤0. Consider the case when t-35\geq 0 and t+2\leq 0.

$$t-35\geq 0 t+2\leq 0$$

This is false for any t.

$$t\in \emptyset$$

Consider the case when t-35\leq 0 and t+2\geq 0.

$$t+2\geq 0 t-35\leq 0$$

The solution satisfying both inequalities is t\in \left[-2,35\right].

$$t\in \begin{bmatrix}-2,35\end{bmatrix}$$

The final solution is the union of the obtained solutions.

$$t\in \begin{bmatrix}-2,35\end{bmatrix}$$