Solve 2x^2-2x-1>0

$$2 x ^ { 2 } - 2 x - 1 > 0$$

Solve for x

$$x\in \left(-\infty,\frac{1-\sqrt{3}}{2}\right)\cup \left(\frac{\sqrt{3}+1}{2},\infty\right)$$

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.

$$2x^{2}-2x-1=0$$

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 2 for a, -2 for b, and -1 for c in the quadratic formula.

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

Do the calculations.

$$x=\frac{2±2\sqrt{3}}{4}$$

Solve the equation x=\frac{2±2\sqrt{3}}{4} when ± is plus and when ± is minus.

$$x=\frac{\sqrt{3}+1}{2} x=\frac{1-\sqrt{3}}{2}$$

Rewrite the inequality by using the obtained solutions.

$$2\left(x-\frac{\sqrt{3}+1}{2}\right)\left(x-\frac{1-\sqrt{3}}{2}\right)>0$$

For the product to be positive, x-\frac{\sqrt{3}+1}{2} and x-\frac{1-\sqrt{3}}{2} have to be both negative or both positive. Consider the case when x-\frac{\sqrt{3}+1}{2} and x-\frac{1-\sqrt{3}}{2} are both negative.

$$x-\frac{\sqrt{3}+1}{2}<0 x-\frac{1-\sqrt{3}}{2}<0$$

The solution satisfying both inequalities is x<\frac{1-\sqrt{3}}{2}.

$$x<\frac{1-\sqrt{3}}{2}$$

Consider the case when x-\frac{\sqrt{3}+1}{2} and x-\frac{1-\sqrt{3}}{2} are both positive.

$$x-\frac{1-\sqrt{3}}{2}>0 x-\frac{\sqrt{3}+1}{2}>0$$

The solution satisfying both inequalities is x>\frac{\sqrt{3}+1}{2}.

$$x>\frac{\sqrt{3}+1}{2}$$

The final solution is the union of the obtained solutions.

$$x<\frac{1-\sqrt{3}}{2}\text{; }x>\frac{\sqrt{3}+1}{2}$$