Solve x^2-2x-4

$${ x }^{ 2 } -2x-4$$

Factor

$$\left(x-\left(1-\sqrt{5}\right)\right)\left(x-\left(\sqrt{5}+1\right)\right)$$

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.

$$x^{2}-2x-4=0$$

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.

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

Square -2.

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

Multiply -4 times -4.

$$x=\frac{-\left(-2\right)±\sqrt{4+16}}{2}$$

Add 4 to 16.

$$x=\frac{-\left(-2\right)±\sqrt{20}}{2}$$

Take the square root of 20.

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

The opposite of -2 is 2.

$$x=\frac{2±2\sqrt{5}}{2}$$

Now solve the equation x=\frac{2±2\sqrt{5}}{2} when ± is plus. Add 2 to 2\sqrt{5}.

$$x=\frac{2\sqrt{5}+2}{2}$$

Divide 2+2\sqrt{5} by 2.

$$x=\sqrt{5}+1$$

Now solve the equation x=\frac{2±2\sqrt{5}}{2} when ± is minus. Subtract 2\sqrt{5} from 2.

$$x=\frac{2-2\sqrt{5}}{2}$$

Divide 2-2\sqrt{5} by 2.

$$x=1-\sqrt{5}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 1+\sqrt{5} for x_{1} and 1-\sqrt{5} for x_{2}.

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

Evaluate

$$x^{2}-2x-4$$