Solve n^2+90n-500>0

$$n ^ { 2 } + 90 n - 500 > 0$$

Solve for n

$$n\in \left(-\infty,-5\sqrt{101}-45\right)\cup \left(5\sqrt{101}-45,\infty\right)$$

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.

$$n^{2}+90n-500=0$$

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, 90 for b, and -500 for c in the quadratic formula.

$$n=\frac{-90±\sqrt{90^{2}-4\times 1\left(-500\right)}}{2}$$

Do the calculations.

$$n=\frac{-90±10\sqrt{101}}{2}$$

Solve the equation n=\frac{-90±10\sqrt{101}}{2} when ± is plus and when ± is minus.

$$n=5\sqrt{101}-45 n=-5\sqrt{101}-45$$

Rewrite the inequality by using the obtained solutions.

$$\left(n-\left(5\sqrt{101}-45\right)\right)\left(n-\left(-5\sqrt{101}-45\right)\right)>0$$

For the product to be positive, n-\left(5\sqrt{101}-45\right) and n-\left(-5\sqrt{101}-45\right) have to be both negative or both positive. Consider the case when n-\left(5\sqrt{101}-45\right) and n-\left(-5\sqrt{101}-45\right) are both negative.

$$n-\left(5\sqrt{101}-45\right)<0 n-\left(-5\sqrt{101}-45\right)<0$$

The solution satisfying both inequalities is n<-5\sqrt{101}-45.

$$n<-5\sqrt{101}-45$$

Consider the case when n-\left(5\sqrt{101}-45\right) and n-\left(-5\sqrt{101}-45\right) are both positive.

$$n-\left(-5\sqrt{101}-45\right)>0 n-\left(5\sqrt{101}-45\right)>0$$

The solution satisfying both inequalities is n>5\sqrt{101}-45.

$$n>5\sqrt{101}-45$$

The final solution is the union of the obtained solutions.

$$n<-5\sqrt{101}-45\text{; }n>5\sqrt{101}-45$$