Solve n^2+16n-200>0

$$n ^ { 2 } + 16 n - 200 > 0$$

Solve for n

$$n\in \left(-\infty,-2\sqrt{66}-8\right)\cup \left(2\sqrt{66}-8,\infty\right)$$

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.

$$n^{2}+16n-200=0$$

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, 16 for b, and -200 for c in the quadratic formula.

$$n=\frac{-16±\sqrt{16^{2}-4\times 1\left(-200\right)}}{2}$$

Do the calculations.

$$n=\frac{-16±4\sqrt{66}}{2}$$

Solve the equation n=\frac{-16±4\sqrt{66}}{2} when ± is plus and when ± is minus.

$$n=2\sqrt{66}-8 n=-2\sqrt{66}-8$$

Rewrite the inequality by using the obtained solutions.

$$\left(n-\left(2\sqrt{66}-8\right)\right)\left(n-\left(-2\sqrt{66}-8\right)\right)>0$$

For the product to be positive, n-\left(2\sqrt{66}-8\right) and n-\left(-2\sqrt{66}-8\right) have to be both negative or both positive. Consider the case when n-\left(2\sqrt{66}-8\right) and n-\left(-2\sqrt{66}-8\right) are both negative.

$$n-\left(2\sqrt{66}-8\right)<0 n-\left(-2\sqrt{66}-8\right)<0$$

The solution satisfying both inequalities is n<-2\sqrt{66}-8.

$$n<-2\sqrt{66}-8$$

Consider the case when n-\left(2\sqrt{66}-8\right) and n-\left(-2\sqrt{66}-8\right) are both positive.

$$n-\left(-2\sqrt{66}-8\right)>0 n-\left(2\sqrt{66}-8\right)>0$$

The solution satisfying both inequalities is n>2\sqrt{66}-8.

$$n>2\sqrt{66}-8$$

The final solution is the union of the obtained solutions.

$$n<-2\sqrt{66}-8\text{; }n>2\sqrt{66}-8$$