Solve x^2-20x-100

$${ x }^{ 2 } -20x-100$$

Factor

$$\left(x-\left(10-10\sqrt{2}\right)\right)\left(x-\left(10\sqrt{2}+10\right)\right)$$

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.

$$x^{2}-20x-100=0$$

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.

$$x=\frac{-\left(-20\right)±\sqrt{\left(-20\right)^{2}-4\left(-100\right)}}{2}$$

Square -20.

$$x=\frac{-\left(-20\right)±\sqrt{400-4\left(-100\right)}}{2}$$

Multiply -4 times -100.

$$x=\frac{-\left(-20\right)±\sqrt{400+400}}{2}$$

Add 400 to 400.

$$x=\frac{-\left(-20\right)±\sqrt{800}}{2}$$

Take the square root of 800.

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

The opposite of -20 is 20.

$$x=\frac{20±20\sqrt{2}}{2}$$

Now solve the equation x=\frac{20±20\sqrt{2}}{2} when ± is plus. Add 20 to 20\sqrt{2}.

$$x=\frac{20\sqrt{2}+20}{2}$$

Divide 20+20\sqrt{2} by 2.

$$x=10\sqrt{2}+10$$

Now solve the equation x=\frac{20±20\sqrt{2}}{2} when ± is minus. Subtract 20\sqrt{2} from 20.

$$x=\frac{20-20\sqrt{2}}{2}$$

Divide 20-20\sqrt{2} by 2.

$$x=10-10\sqrt{2}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 10+10\sqrt{2} for x_{1} and 10-10\sqrt{2} for x_{2}.

$$x^{2}-20x-100=\left(x-\left(10\sqrt{2}+10\right)\right)\left(x-\left(10-10\sqrt{2}\right)\right)$$

Evaluate

$$x^{2}-20x-100$$