Solve x^2+25x+5=

$${ x }^{ 2 } +25x+5=$$

Factor

$$\left(x-\frac{-11\sqrt{5}-25}{2}\right)\left(x-\frac{11\sqrt{5}-25}{2}\right)$$

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.

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

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.

$$x=\frac{-25±\sqrt{25^{2}-4\times 5}}{2}$$

Square 25.

$$x=\frac{-25±\sqrt{625-4\times 5}}{2}$$

Multiply -4 times 5.

$$x=\frac{-25±\sqrt{625-20}}{2}$$

Add 625 to -20.

$$x=\frac{-25±\sqrt{605}}{2}$$

Take the square root of 605.

$$x=\frac{-25±11\sqrt{5}}{2}$$

Now solve the equation x=\frac{-25±11\sqrt{5}}{2} when ± is plus. Add -25 to 11\sqrt{5}.

$$x=\frac{11\sqrt{5}-25}{2}$$

Now solve the equation x=\frac{-25±11\sqrt{5}}{2} when ± is minus. Subtract 11\sqrt{5} from -25.

$$x=\frac{-11\sqrt{5}-25}{2}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{-25+11\sqrt{5}}{2} for x_{1} and \frac{-25-11\sqrt{5}}{2} for x_{2}.

$$x^{2}+25x+5=\left(x-\frac{11\sqrt{5}-25}{2}\right)\left(x-\frac{-11\sqrt{5}-25}{2}\right)$$

Evaluate

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