$$ \begin{aligned} x\cdot2-5x+6 &= (x-2)(x-3)&& \text{simplify left and right hand side} \\[1 em]x\cdot2-5x+6 &= x^2-3x-2x+6&& \\[1 em]-3x+6 &= x^2-5x+6&& \text{move all terms to the left hand side } \\[1 em]-3x+6-x^2+5x-6 &= 0&& \text{simplify left side} \\[1 em]-3x+6-x^2+5x-6 &= 0&& \\[1 em]-x^2+2x &= 0&& \\[1 em] \end{aligned} $$
In order to solve $ \color{blue}{ -x^{2}+2x = 0 } $, first we need to factor our $ x $.
$$ -x^{2}+2x = x \left( -x+2 \right) $$
$ x = 0 $ is a root of multiplicity $ 1 $.
The second root can be found by solving equation $ -x+2 = 0$.
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