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 $ax^{2}+bx+c=0$.
$$5y^{2}-6y=0$$
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.
Now solve the equation $y=\frac{6±6}{10}$ when $±$ is plus. Add $6$ to $6$.
$$y=\frac{12}{10}$$
Reduce the fraction $\frac{12}{10}$ to lowest terms by extracting and canceling out $2$.
$$y=\frac{6}{5}$$
Now solve the equation $y=\frac{6±6}{10}$ when $±$ is minus. Subtract $6$ from $6$.
$$y=\frac{0}{10}$$
Divide $0$ by $10$.
$$y=0$$
Factor the original expression using $ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$. Substitute $\frac{6}{5}$ for $x_{1}$ and $0$ for $x_{2}$.
$$5y^{2}-6y=5\left(y-\frac{6}{5}\right)y$$
Subtract $\frac{6}{5}$ from $y$ by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.
$$5y^{2}-6y=5\times \frac{5y-6}{5}y$$
Cancel out $5$, the greatest common factor in $5$ and $5$.
