Solve 0>n^2-5n-8

$$0 > { n }^{ 2 } -5n-8$$

Solve for n

$$n\in \left(\frac{5-\sqrt{57}}{2},\frac{\sqrt{57}+5}{2}\right)$$

Steps Using the Quadratic Formula

Swap sides so that all variable terms are on the left hand side. This changes the sign direction.

$$n^{2}-5n-8<0$$

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.

$$n^{2}-5n-8=0$$

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, -5 for b, and -8 for c in the quadratic formula.

$$n=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 1\left(-8\right)}}{2}$$

Do the calculations.

$$n=\frac{5±\sqrt{57}}{2}$$

Solve the equation n=\frac{5±\sqrt{57}}{2} when ± is plus and when ± is minus.

$$n=\frac{\sqrt{57}+5}{2} n=\frac{5-\sqrt{57}}{2}$$

Rewrite the inequality by using the obtained solutions.

$$\left(n-\frac{\sqrt{57}+5}{2}\right)\left(n-\frac{5-\sqrt{57}}{2}\right)<0$$

For the product to be negative, n-\frac{\sqrt{57}+5}{2} and n-\frac{5-\sqrt{57}}{2} have to be of the opposite signs. Consider the case when n-\frac{\sqrt{57}+5}{2} is positive and n-\frac{5-\sqrt{57}}{2} is negative.

$$n-\frac{\sqrt{57}+5}{2}>0 n-\frac{5-\sqrt{57}}{2}<0$$

This is false for any n.

$$n\in \emptyset$$

Consider the case when n-\frac{5-\sqrt{57}}{2} is positive and n-\frac{\sqrt{57}+5}{2} is negative.

$$n-\frac{5-\sqrt{57}}{2}>0 n-\frac{\sqrt{57}+5}{2}<0$$

The solution satisfying both inequalities is n\in \left(\frac{5-\sqrt{57}}{2},\frac{\sqrt{57}+5}{2}\right).

$$n\in \left(\frac{5-\sqrt{57}}{2},\frac{\sqrt{57}+5}{2}\right)$$

The final solution is the union of the obtained solutions.

$$n\in \left(\frac{5-\sqrt{57}}{2},\frac{\sqrt{57}+5}{2}\right)$$