Solve x^2+25x-50

$${ x }^{ 2 } +25x-50$$

Factor

$$\left(x-\frac{-5\sqrt{33}-25}{2}\right)\left(x-\frac{5\sqrt{33}-25}{2}\right)$$

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.

$$x^{2}+25x-50=0$$

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.

$$x=\frac{-25±\sqrt{25^{2}-4\left(-50\right)}}{2}$$

Square 25.

$$x=\frac{-25±\sqrt{625-4\left(-50\right)}}{2}$$

Multiply -4 times -50.

$$x=\frac{-25±\sqrt{625+200}}{2}$$

Add 625 to 200.

$$x=\frac{-25±\sqrt{825}}{2}$$

Take the square root of 825.

$$x=\frac{-25±5\sqrt{33}}{2}$$

Now solve the equation x=\frac{-25±5\sqrt{33}}{2} when ± is plus. Add -25 to 5\sqrt{33}.

$$x=\frac{5\sqrt{33}-25}{2}$$

Now solve the equation x=\frac{-25±5\sqrt{33}}{2} when ± is minus. Subtract 5\sqrt{33} from -25.

$$x=\frac{-5\sqrt{33}-25}{2}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{-25+5\sqrt{33}}{2} for x_{1} and \frac{-25-5\sqrt{33}}{2} for x_{2}.

$$x^{2}+25x-50=\left(x-\frac{5\sqrt{33}-25}{2}\right)\left(x-\frac{-5\sqrt{33}-25}{2}\right)$$

Evaluate

$$x^{2}+25x-50$$