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 $ax^{2}+bx+c=0$.
$$x^{2}+10x+5=0$$
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.
$$x=\frac{-10±\sqrt{10^{2}-4\times 5}}{2}$$
Square $10$.
$$x=\frac{-10±\sqrt{100-4\times 5}}{2}$$
Multiply $-4$ times $5$.
$$x=\frac{-10±\sqrt{100-20}}{2}$$
Add $100$ to $-20$.
$$x=\frac{-10±\sqrt{80}}{2}$$
Take the square root of $80$.
$$x=\frac{-10±4\sqrt{5}}{2}$$
Now solve the equation $x=\frac{-10±4\sqrt{5}}{2}$ when $±$ is plus. Add $-10$ to $4\sqrt{5}$.
$$x=\frac{4\sqrt{5}-10}{2}$$
Divide $-10+4\sqrt{5}$ by $2$.
$$x=2\sqrt{5}-5$$
Now solve the equation $x=\frac{-10±4\sqrt{5}}{2}$ when $±$ is minus. Subtract $4\sqrt{5}$ from $-10$.
$$x=\frac{-4\sqrt{5}-10}{2}$$
Divide $-10-4\sqrt{5}$ by $2$.
$$x=-2\sqrt{5}-5$$
Factor the original expression using $ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$. Substitute $-5+2\sqrt{5}$ for $x_{1}$ and $-5-2\sqrt{5}$ for $x_{2}$.
