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 $ax^{2}+bx+c=0$.
$$y^{2}+3y-20=0$$
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.
$$y=\frac{-3±\sqrt{3^{2}-4\left(-20\right)}}{2}$$
Square $3$.
$$y=\frac{-3±\sqrt{9-4\left(-20\right)}}{2}$$
Multiply $-4$ times $-20$.
$$y=\frac{-3±\sqrt{9+80}}{2}$$
Add $9$ to $80$.
$$y=\frac{-3±\sqrt{89}}{2}$$
Now solve the equation $y=\frac{-3±\sqrt{89}}{2}$ when $±$ is plus. Add $-3$ to $\sqrt{89}$.
$$y=\frac{\sqrt{89}-3}{2}$$
Now solve the equation $y=\frac{-3±\sqrt{89}}{2}$ when $±$ is minus. Subtract $\sqrt{89}$ from $-3$.
$$y=\frac{-\sqrt{89}-3}{2}$$
Factor the original expression using $ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$. Substitute $\frac{-3+\sqrt{89}}{2}$ for $x_{1}$ and $\frac{-3-\sqrt{89}}{2}$ for $x_{2}$.
