$$ \begin{aligned} \frac{1}{6}x^2-\frac{3}{2}x &= 0&& \text{multiply ALL terms by } \color{blue}{ 6 }. \\[1 em]6 \cdot \frac{1}{6}x^2-6\frac{3}{2}x &= 6\cdot0&& \text{cancel out the denominators} \\[1 em]x^2-9x &= 0&& \\[1 em] \end{aligned} $$
In order to solve $ \color{blue}{ x^{2}-9x = 0 } $, first we need to factor our $ x $.
$$ x^{2}-9x = x \left( x-9 \right) $$
$ x = 0 $ is a root of multiplicity $ 1 $.
The second root can be found by solving equation $ x-9 = 0$.
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