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 $ax^{2}+bx+c=0$.
$$4y^{2}+9y+1=0$$
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.
$$y=\frac{-9±\sqrt{9^{2}-4\times 4}}{2\times 4}$$
Square $9$.
$$y=\frac{-9±\sqrt{81-4\times 4}}{2\times 4}$$
Multiply $-4$ times $4$.
$$y=\frac{-9±\sqrt{81-16}}{2\times 4}$$
Add $81$ to $-16$.
$$y=\frac{-9±\sqrt{65}}{2\times 4}$$
Multiply $2$ times $4$.
$$y=\frac{-9±\sqrt{65}}{8}$$
Now solve the equation $y=\frac{-9±\sqrt{65}}{8}$ when $±$ is plus. Add $-9$ to $\sqrt{65}$.
$$y=\frac{\sqrt{65}-9}{8}$$
Now solve the equation $y=\frac{-9±\sqrt{65}}{8}$ when $±$ is minus. Subtract $\sqrt{65}$ from $-9$.
$$y=\frac{-\sqrt{65}-9}{8}$$
Factor the original expression using $ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$. Substitute $\frac{-9+\sqrt{65}}{8}$ for $x_{1}$ and $\frac{-9-\sqrt{65}}{8}$ for $x_{2}$.
