Solve x^2-2x-5<0

$$x ^ { 2 } - 2 x - 5 < 0$$

Solve for x

$$x\in \left(1-\sqrt{6},\sqrt{6}+1\right)$$

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.

$$x^{2}-2x-5=0$$

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, -2 for b, and -5 for c in the quadratic formula.

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

Do the calculations.

$$x=\frac{2±2\sqrt{6}}{2}$$

Solve the equation x=\frac{2±2\sqrt{6}}{2} when ± is plus and when ± is minus.

$$x=\sqrt{6}+1 x=1-\sqrt{6}$$

Rewrite the inequality by using the obtained solutions.

$$\left(x-\left(\sqrt{6}+1\right)\right)\left(x-\left(1-\sqrt{6}\right)\right)<0$$

For the product to be negative, x-\left(\sqrt{6}+1\right) and x-\left(1-\sqrt{6}\right) have to be of the opposite signs. Consider the case when x-\left(\sqrt{6}+1\right) is positive and x-\left(1-\sqrt{6}\right) is negative.

$$x-\left(\sqrt{6}+1\right)>0 x-\left(1-\sqrt{6}\right)<0$$

This is false for any x.

$$x\in \emptyset$$

Consider the case when x-\left(1-\sqrt{6}\right) is positive and x-\left(\sqrt{6}+1\right) is negative.

$$x-\left(1-\sqrt{6}\right)>0 x-\left(\sqrt{6}+1\right)<0$$

The solution satisfying both inequalities is x\in \left(1-\sqrt{6},\sqrt{6}+1\right).

$$x\in \left(1-\sqrt{6},\sqrt{6}+1\right)$$

The final solution is the union of the obtained solutions.

$$x\in \left(1-\sqrt{6},\sqrt{6}+1\right)$$