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 $ax^{2}+bx+c=0$.
$$5x^{2}+25x=0$$
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.
$$x=\frac{-25±\sqrt{25^{2}}}{2\times 5}$$
Take the square root of $25^{2}$.
$$x=\frac{-25±25}{2\times 5}$$
Multiply $2$ times $5$.
$$x=\frac{-25±25}{10}$$
Now solve the equation $x=\frac{-25±25}{10}$ when $±$ is plus. Add $-25$ to $25$.
$$x=\frac{0}{10}$$
Divide $0$ by $10$.
$$x=0$$
Now solve the equation $x=\frac{-25±25}{10}$ when $±$ is minus. Subtract $25$ from $-25$.
$$x=-\frac{50}{10}$$
Divide $-50$ by $10$.
$$x=-5$$
Factor the original expression using $ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$. Substitute $0$ for $x_{1}$ and $-5$ for $x_{2}$.
$$5x^{2}+25x=5x\left(x-\left(-5\right)\right)$$
Simplify all the expressions of the form $p-\left(-q\right)$ to $p+q$.
