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 $ax^{2}+bx+c=0$.
$$9x^{2}-10x=0$$
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 $x=\frac{10±10}{18}$ when $±$ is plus. Add $10$ to $10$.
$$x=\frac{20}{18}$$
Reduce the fraction $\frac{20}{18}$ to lowest terms by extracting and canceling out $2$.
$$x=\frac{10}{9}$$
Now solve the equation $x=\frac{10±10}{18}$ when $±$ is minus. Subtract $10$ from $10$.
$$x=\frac{0}{18}$$
Divide $0$ by $18$.
$$x=0$$
Factor the original expression using $ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$. Substitute $\frac{10}{9}$ for $x_{1}$ and $0$ for $x_{2}$.
$$9x^{2}-10x=9\left(x-\frac{10}{9}\right)x$$
Subtract $\frac{10}{9}$ from $x$ by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.
$$9x^{2}-10x=9\times \frac{9x-10}{9}x$$
Cancel out $9$, the greatest common factor in $9$ and $9$.
