Solve 5x^2+25x+5

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

Factor

$$5\left(x-\frac{-\sqrt{21}-5}{2}\right)\left(x-\frac{\sqrt{21}-5}{2}\right)$$

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.

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

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.

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

Square 25.

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

Multiply -4 times 5.

$$x=\frac{-25±\sqrt{625-20\times 5}}{2\times 5}$$

Multiply -20 times 5.

$$x=\frac{-25±\sqrt{625-100}}{2\times 5}$$

Add 625 to -100.

$$x=\frac{-25±\sqrt{525}}{2\times 5}$$

Take the square root of 525.

$$x=\frac{-25±5\sqrt{21}}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{-25±5\sqrt{21}}{10}$$

Now solve the equation x=\frac{-25±5\sqrt{21}}{10} when ± is plus. Add -25 to 5\sqrt{21}.

$$x=\frac{5\sqrt{21}-25}{10}$$

Divide -25+5\sqrt{21} by 10.

$$x=\frac{\sqrt{21}-5}{2}$$

Now solve the equation x=\frac{-25±5\sqrt{21}}{10} when ± is minus. Subtract 5\sqrt{21} from -25.

$$x=\frac{-5\sqrt{21}-25}{10}$$

Divide -25-5\sqrt{21} by 10.

$$x=\frac{-\sqrt{21}-5}{2}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{-5+\sqrt{21}}{2} for x_{1} and \frac{-5-\sqrt{21}}{2} for x_{2}.

$$5x^{2}+25x+5=5\left(x-\frac{\sqrt{21}-5}{2}\right)\left(x-\frac{-\sqrt{21}-5}{2}\right)$$

Evaluate

$$5\left(x^{2}+5x+1\right)$$