Solve n^2+n-400leq0

$$n ^ { 2 } + n - 400 \leq 0$$

Solve for n

$$n\in \begin{bmatrix}\frac{-\sqrt{1601}-1}{2},\frac{\sqrt{1601}-1}{2}\end{bmatrix}$$

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.

$$n^{2}+n-400=0$$

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, 1 for b, and -400 for c in the quadratic formula.

$$n=\frac{-1±\sqrt{1^{2}-4\times 1\left(-400\right)}}{2}$$

Do the calculations.

$$n=\frac{-1±\sqrt{1601}}{2}$$

Solve the equation n=\frac{-1±\sqrt{1601}}{2} when ± is plus and when ± is minus.

$$n=\frac{\sqrt{1601}-1}{2} n=\frac{-\sqrt{1601}-1}{2}$$

Rewrite the inequality by using the obtained solutions.

$$\left(n-\frac{\sqrt{1601}-1}{2}\right)\left(n-\frac{-\sqrt{1601}-1}{2}\right)\leq 0$$

For the product to be ≤0, one of the values n-\frac{\sqrt{1601}-1}{2} and n-\frac{-\sqrt{1601}-1}{2} has to be ≥0 and the other has to be ≤0. Consider the case when n-\frac{\sqrt{1601}-1}{2}\geq 0 and n-\frac{-\sqrt{1601}-1}{2}\leq 0.

$$n-\frac{\sqrt{1601}-1}{2}\geq 0 n-\frac{-\sqrt{1601}-1}{2}\leq 0$$

This is false for any n.

$$n\in \emptyset$$

Consider the case when n-\frac{\sqrt{1601}-1}{2}\leq 0 and n-\frac{-\sqrt{1601}-1}{2}\geq 0.

$$n-\frac{-\sqrt{1601}-1}{2}\geq 0 n-\frac{\sqrt{1601}-1}{2}\leq 0$$

The solution satisfying both inequalities is n\in \left[\frac{-\sqrt{1601}-1}{2},\frac{\sqrt{1601}-1}{2}\right].

$$n\in \begin{bmatrix}\frac{-\sqrt{1601}-1}{2},\frac{\sqrt{1601}-1}{2}\end{bmatrix}$$

The final solution is the union of the obtained solutions.

$$n\in \begin{bmatrix}\frac{-\sqrt{1601}-1}{2},\frac{\sqrt{1601}-1}{2}\end{bmatrix}$$