Solve (x^2-2x-2)

$$( x ^ { 2 } - 2 x - 2 )$$

Factor

$$\left(x-\left(1-\sqrt{3}\right)\right)\left(x-\left(\sqrt{3}+1\right)\right)$$

Steps Using the Quadratic Formula

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.

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

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

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

Square -2.

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

Multiply -4 times -2.

$$x=\frac{-\left(-2\right)±\sqrt{4+8}}{2}$$

Add 4 to 8.

$$x=\frac{-\left(-2\right)±\sqrt{12}}{2}$$

Take the square root of 12.

$$x=\frac{-\left(-2\right)±2\sqrt{3}}{2}$$

The opposite of -2 is 2.

$$x=\frac{2±2\sqrt{3}}{2}$$

Now solve the equation x=\frac{2±2\sqrt{3}}{2} when ± is plus. Add 2 to 2\sqrt{3}.

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

Divide 2+2\sqrt{3} by 2.

$$x=\sqrt{3}+1$$

Now solve the equation x=\frac{2±2\sqrt{3}}{2} when ± is minus. Subtract 2\sqrt{3} from 2.

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

Divide 2-2\sqrt{3} by 2.

$$x=1-\sqrt{3}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 1+\sqrt{3} for x_{1} and 1-\sqrt{3} for x_{2}.

$$x^{2}-2x-2=\left(x-\left(\sqrt{3}+1\right)\right)\left(x-\left(1-\sqrt{3}\right)\right)$$

Evaluate

$$x^{2}-2x-2$$