Solve 3n^2-3n-199leq0

$$3 n ^ { 2 } - 3 n - 199 \leq 0$$

Solve for n

$$n\in \begin{bmatrix}-\frac{\sqrt{2397}}{6}+\frac{1}{2},\frac{\sqrt{2397}}{6}+\frac{1}{2}\end{bmatrix}$$

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.

$$3n^{2}-3n-199=0$$

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 3 for a, -3 for b, and -199 for c in the quadratic formula.

$$n=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\times 3\left(-199\right)}}{2\times 3}$$

Do the calculations.

$$n=\frac{3±\sqrt{2397}}{6}$$

Solve the equation n=\frac{3±\sqrt{2397}}{6} when ± is plus and when ± is minus.

$$n=\frac{\sqrt{2397}}{6}+\frac{1}{2} n=-\frac{\sqrt{2397}}{6}+\frac{1}{2}$$

Rewrite the inequality by using the obtained solutions.

$$3\left(n-\left(\frac{\sqrt{2397}}{6}+\frac{1}{2}\right)\right)\left(n-\left(-\frac{\sqrt{2397}}{6}+\frac{1}{2}\right)\right)\leq 0$$

For the product to be ≤0, one of the values n-\left(\frac{\sqrt{2397}}{6}+\frac{1}{2}\right) and n-\left(-\frac{\sqrt{2397}}{6}+\frac{1}{2}\right) has to be ≥0 and the other has to be ≤0. Consider the case when n-\left(\frac{\sqrt{2397}}{6}+\frac{1}{2}\right)\geq 0 and n-\left(-\frac{\sqrt{2397}}{6}+\frac{1}{2}\right)\leq 0.

$$n-\left(\frac{\sqrt{2397}}{6}+\frac{1}{2}\right)\geq 0 n-\left(-\frac{\sqrt{2397}}{6}+\frac{1}{2}\right)\leq 0$$

This is false for any n.

$$n\in \emptyset$$

Consider the case when n-\left(\frac{\sqrt{2397}}{6}+\frac{1}{2}\right)\leq 0 and n-\left(-\frac{\sqrt{2397}}{6}+\frac{1}{2}\right)\geq 0.

$$n-\left(-\frac{\sqrt{2397}}{6}+\frac{1}{2}\right)\geq 0 n-\left(\frac{\sqrt{2397}}{6}+\frac{1}{2}\right)\leq 0$$

The solution satisfying both inequalities is n\in \left[-\frac{\sqrt{2397}}{6}+\frac{1}{2},\frac{\sqrt{2397}}{6}+\frac{1}{2}\right].

$$n\in \begin{bmatrix}-\frac{\sqrt{2397}}{6}+\frac{1}{2},\frac{\sqrt{2397}}{6}+\frac{1}{2}\end{bmatrix}$$

The final solution is the union of the obtained solutions.

$$n\in \begin{bmatrix}-\frac{\sqrt{2397}}{6}+\frac{1}{2},\frac{\sqrt{2397}}{6}+\frac{1}{2}\end{bmatrix}$$