Solve x^2-20x-25

$${ x }^{ 2 } -20x-25$$

Factor

$$\left(x-\left(10-5\sqrt{5}\right)\right)\left(x-\left(5\sqrt{5}+10\right)\right)$$

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.

$$x^{2}-20x-25=0$$

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.

$$x=\frac{-\left(-20\right)±\sqrt{\left(-20\right)^{2}-4\left(-25\right)}}{2}$$

Square -20.

$$x=\frac{-\left(-20\right)±\sqrt{400-4\left(-25\right)}}{2}$$

Multiply -4 times -25.

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

Add 400 to 100.

$$x=\frac{-\left(-20\right)±\sqrt{500}}{2}$$

Take the square root of 500.

$$x=\frac{-\left(-20\right)±10\sqrt{5}}{2}$$

The opposite of -20 is 20.

$$x=\frac{20±10\sqrt{5}}{2}$$

Now solve the equation x=\frac{20±10\sqrt{5}}{2} when ± is plus. Add 20 to 10\sqrt{5}.

$$x=\frac{10\sqrt{5}+20}{2}$$

Divide 20+10\sqrt{5} by 2.

$$x=5\sqrt{5}+10$$

Now solve the equation x=\frac{20±10\sqrt{5}}{2} when ± is minus. Subtract 10\sqrt{5} from 20.

$$x=\frac{20-10\sqrt{5}}{2}$$

Divide 20-10\sqrt{5} by 2.

$$x=10-5\sqrt{5}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 10+5\sqrt{5} for x_{1} and 10-5\sqrt{5} for x_{2}.

$$x^{2}-20x-25=\left(x-\left(5\sqrt{5}+10\right)\right)\left(x-\left(10-5\sqrt{5}\right)\right)$$

Evaluate

$$x^{2}-20x-25$$