$$-x^2+\frac{20}{10}x = 0$$
Answer
$$ \begin{matrix}x_1 = 0 & x_2 = 2 \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} -x^2+\frac{20}{10}x &= 0&& \text{multiply ALL terms by } \color{blue}{ 10 }. \\[1 em]-10x^2+10 \cdot \frac{20}{10}x &= 10\cdot0&& \text{cancel out the denominators} \\[1 em]-10x^2+20x &= 0&& \\[1 em] \end{aligned} $$
In order to solve $ \color{blue}{ -10x^{2}+20x = 0 } $, first we need to factor our $ x $.
$$ -10x^{2}+20x = x \left( -10x+20 \right) $$$ x = 0 $ is a root of multiplicity $ 1 $.
The second root can be found by solving equation $ -10x+20 = 0$.
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