Solve n^2+6n-600leq0

$$n ^ { 2 } + 6 n - 600 \leq 0$$

Solve for n

$$n\in \begin{bmatrix}-\sqrt{609}-3,\sqrt{609}-3\end{bmatrix}$$

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.

$$n^{2}+6n-600=0$$

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, 6 for b, and -600 for c in the quadratic formula.

$$n=\frac{-6±\sqrt{6^{2}-4\times 1\left(-600\right)}}{2}$$

Do the calculations.

$$n=\frac{-6±2\sqrt{609}}{2}$$

Solve the equation n=\frac{-6±2\sqrt{609}}{2} when ± is plus and when ± is minus.

$$n=\sqrt{609}-3 n=-\sqrt{609}-3$$

Rewrite the inequality by using the obtained solutions.

$$\left(n-\left(\sqrt{609}-3\right)\right)\left(n-\left(-\sqrt{609}-3\right)\right)\leq 0$$

For the product to be ≤0, one of the values n-\left(\sqrt{609}-3\right) and n-\left(-\sqrt{609}-3\right) has to be ≥0 and the other has to be ≤0. Consider the case when n-\left(\sqrt{609}-3\right)\geq 0 and n-\left(-\sqrt{609}-3\right)\leq 0.

$$n-\left(\sqrt{609}-3\right)\geq 0 n-\left(-\sqrt{609}-3\right)\leq 0$$

This is false for any n.

$$n\in \emptyset$$

Consider the case when n-\left(\sqrt{609}-3\right)\leq 0 and n-\left(-\sqrt{609}-3\right)\geq 0.

$$n-\left(-\sqrt{609}-3\right)\geq 0 n-\left(\sqrt{609}-3\right)\leq 0$$

The solution satisfying both inequalities is n\in \left[-\left(\sqrt{609}+3\right),\sqrt{609}-3\right].

$$n\in \begin{bmatrix}-\left(\sqrt{609}+3\right),\sqrt{609}-3\end{bmatrix}$$

The final solution is the union of the obtained solutions.

$$n\in \begin{bmatrix}-\sqrt{609}-3,\sqrt{609}-3\end{bmatrix}$$