Solve n^2-2n-8leq0

$$n ^ { 2 } - 2 n - 8 \leq 0$$

Solve for n

$$n\in \begin{bmatrix}-2,4\end{bmatrix}$$

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.

$$n^{2}-2n-8=0$$

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, -2 for b, and -8 for c in the quadratic formula.

$$n=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\times 1\left(-8\right)}}{2}$$

Do the calculations.

$$n=\frac{2±6}{2}$$

Solve the equation n=\frac{2±6}{2} when ± is plus and when ± is minus.

$$n=4 n=-2$$

Rewrite the inequality by using the obtained solutions.

$$\left(n-4\right)\left(n+2\right)\leq 0$$

For the product to be ≤0, one of the values n-4 and n+2 has to be ≥0 and the other has to be ≤0. Consider the case when n-4\geq 0 and n+2\leq 0.

$$n-4\geq 0 n+2\leq 0$$

This is false for any n.

$$n\in \emptyset$$

Consider the case when n-4\leq 0 and n+2\geq 0.

$$n+2\geq 0 n-4\leq 0$$

The solution satisfying both inequalities is n\in \left[-2,4\right].

$$n\in \begin{bmatrix}-2,4\end{bmatrix}$$

The final solution is the union of the obtained solutions.

$$n\in \begin{bmatrix}-2,4\end{bmatrix}$$