Solve x^2-20x+40

$${ x }^{ 2 } -20x+40$$

Factor

$$\left(x-\left(10-2\sqrt{15}\right)\right)\left(x-\left(2\sqrt{15}+10\right)\right)$$

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.

$$x^{2}-20x+40=0$$

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.

$$x=\frac{-\left(-20\right)±\sqrt{\left(-20\right)^{2}-4\times 40}}{2}$$

Square -20.

$$x=\frac{-\left(-20\right)±\sqrt{400-4\times 40}}{2}$$

Multiply -4 times 40.

$$x=\frac{-\left(-20\right)±\sqrt{400-160}}{2}$$

Add 400 to -160.

$$x=\frac{-\left(-20\right)±\sqrt{240}}{2}$$

Take the square root of 240.

$$x=\frac{-\left(-20\right)±4\sqrt{15}}{2}$$

The opposite of -20 is 20.

$$x=\frac{20±4\sqrt{15}}{2}$$

Now solve the equation x=\frac{20±4\sqrt{15}}{2} when ± is plus. Add 20 to 4\sqrt{15}.

$$x=\frac{4\sqrt{15}+20}{2}$$

Divide 20+4\sqrt{15} by 2.

$$x=2\sqrt{15}+10$$

Now solve the equation x=\frac{20±4\sqrt{15}}{2} when ± is minus. Subtract 4\sqrt{15} from 20.

$$x=\frac{20-4\sqrt{15}}{2}$$

Divide 20-4\sqrt{15} by 2.

$$x=10-2\sqrt{15}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 10+2\sqrt{15} for x_{1} and 10-2\sqrt{15} for x_{2}.

$$x^{2}-20x+40=\left(x-\left(2\sqrt{15}+10\right)\right)\left(x-\left(10-2\sqrt{15}\right)\right)$$

Evaluate

$$x^{2}-20x+40$$