Solve n^2-10n-1<0

$${ n }^{ 2 } -10n-1 < 0$$

Solve for n

$$n\in \left(5-\sqrt{26},\sqrt{26}+5\right)$$

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.

$$n^{2}-10n-1=0$$

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 1 for a, -10 for b, and -1 for c in the quadratic formula.

$$n=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\times 1\left(-1\right)}}{2}$$

Do the calculations.

$$n=\frac{10±2\sqrt{26}}{2}$$

Solve the equation n=\frac{10±2\sqrt{26}}{2} when ± is plus and when ± is minus.

$$n=\sqrt{26}+5 n=5-\sqrt{26}$$

Rewrite the inequality by using the obtained solutions.

$$\left(n-\left(\sqrt{26}+5\right)\right)\left(n-\left(5-\sqrt{26}\right)\right)<0$$

For the product to be negative, n-\left(\sqrt{26}+5\right) and n-\left(5-\sqrt{26}\right) have to be of the opposite signs. Consider the case when n-\left(\sqrt{26}+5\right) is positive and n-\left(5-\sqrt{26}\right) is negative.

$$n-\left(\sqrt{26}+5\right)>0 n-\left(5-\sqrt{26}\right)<0$$

This is false for any n.

$$n\in \emptyset$$

Consider the case when n-\left(5-\sqrt{26}\right) is positive and n-\left(\sqrt{26}+5\right) is negative.

$$n-\left(5-\sqrt{26}\right)>0 n-\left(\sqrt{26}+5\right)<0$$

The solution satisfying both inequalities is n\in \left(5-\sqrt{26},\sqrt{26}+5\right).

$$n\in \left(5-\sqrt{26},\sqrt{26}+5\right)$$

The final solution is the union of the obtained solutions.

$$n\in \left(5-\sqrt{26},\sqrt{26}+5\right)$$