Solve x^2-5x-25

$${ x }^{ 2 } -5x-25$$

Factor

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

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.

$$x^{2}-5x-25=0$$

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.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\left(-25\right)}}{2}$$

Square -5.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\left(-25\right)}}{2}$$

Multiply -4 times -25.

$$x=\frac{-\left(-5\right)±\sqrt{25+100}}{2}$$

Add 25 to 100.

$$x=\frac{-\left(-5\right)±\sqrt{125}}{2}$$

Take the square root of 125.

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

The opposite of -5 is 5.

$$x=\frac{5±5\sqrt{5}}{2}$$

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

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

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

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

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

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

Evaluate

$$x^{2}-5x-25$$