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 $ax^{2}+bx+c=0$.
$$-9y^{2}+15y=0$$
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.
$$y=\frac{-15±\sqrt{15^{2}}}{2\left(-9\right)}$$
Take the square root of $15^{2}$.
$$y=\frac{-15±15}{2\left(-9\right)}$$
Multiply $2$ times $-9$.
$$y=\frac{-15±15}{-18}$$
Now solve the equation $y=\frac{-15±15}{-18}$ when $±$ is plus. Add $-15$ to $15$.
$$y=\frac{0}{-18}$$
Divide $0$ by $-18$.
$$y=0$$
Now solve the equation $y=\frac{-15±15}{-18}$ when $±$ is minus. Subtract $15$ from $-15$.
$$y=-\frac{30}{-18}$$
Reduce the fraction $\frac{-30}{-18}$ to lowest terms by extracting and canceling out $6$.
$$y=\frac{5}{3}$$
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 $\frac{5}{3}$ for $x_{2}$.
$$-9y^{2}+15y=-9y\left(y-\frac{5}{3}\right)$$
Subtract $\frac{5}{3}$ from $y$ by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.
$$-9y^{2}+15y=-9y\times \frac{-3y+5}{-3}$$
Cancel out $3$, the greatest common factor in $-9$ and $-3$.
