$$ \begin{aligned} x^2-x &= 2x&& \text{move all terms to the left hand side } \\[1 em]x^2-x-2x &= 0&& \text{simplify left side} \\[1 em]x^2-3x &= 0&& \\[1 em] \end{aligned} $$
In order to solve $ \color{blue}{ x^{2}-3x = 0 } $, first we need to factor our $ x $.
$$ x^{2}-3x = x \left( x-3 \right) $$
$ x = 0 $ is a root of multiplicity $ 1 $.
The second root can be found by solving equation $ x-3 = 0$.
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