Solve 13n^2-41n-1>0

$$13 n ^ { 2 } - 41 n - 1 > 0$$

Solve for n

$$n\in \left(-\infty,\frac{41-\sqrt{1733}}{26}\right)\cup \left(\frac{\sqrt{1733}+41}{26},\infty\right)$$

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.

$$13n^{2}-41n-1=0$$

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 13 for a, -41 for b, and -1 for c in the quadratic formula.

$$n=\frac{-\left(-41\right)±\sqrt{\left(-41\right)^{2}-4\times 13\left(-1\right)}}{2\times 13}$$

Do the calculations.

$$n=\frac{41±\sqrt{1733}}{26}$$

Solve the equation n=\frac{41±\sqrt{1733}}{26} when ± is plus and when ± is minus.

$$n=\frac{\sqrt{1733}+41}{26} n=\frac{41-\sqrt{1733}}{26}$$

Rewrite the inequality by using the obtained solutions.

$$13\left(n-\frac{\sqrt{1733}+41}{26}\right)\left(n-\frac{41-\sqrt{1733}}{26}\right)>0$$

For the product to be positive, n-\frac{\sqrt{1733}+41}{26} and n-\frac{41-\sqrt{1733}}{26} have to be both negative or both positive. Consider the case when n-\frac{\sqrt{1733}+41}{26} and n-\frac{41-\sqrt{1733}}{26} are both negative.

$$n-\frac{\sqrt{1733}+41}{26}<0 n-\frac{41-\sqrt{1733}}{26}<0$$

The solution satisfying both inequalities is n<\frac{41-\sqrt{1733}}{26}.

$$n<\frac{41-\sqrt{1733}}{26}$$

Consider the case when n-\frac{\sqrt{1733}+41}{26} and n-\frac{41-\sqrt{1733}}{26} are both positive.

$$n-\frac{41-\sqrt{1733}}{26}>0 n-\frac{\sqrt{1733}+41}{26}>0$$

The solution satisfying both inequalities is n>\frac{\sqrt{1733}+41}{26}.

$$n>\frac{\sqrt{1733}+41}{26}$$

The final solution is the union of the obtained solutions.

$$n<\frac{41-\sqrt{1733}}{26}\text{; }n>\frac{\sqrt{1733}+41}{26}$$