Question
$$4 \cdot \frac{x}{3} = 2 \cdot \frac{x^2}{5}$$
Answer
$$ \begin{matrix}x_1 = 0 & x_2 = \dfrac{ 10 }{ 3 } \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} 4 \cdot \frac{x}{3} &= 2 \cdot \frac{x^2}{5}&& \text{multiply ALL terms by } \color{blue}{ 15 }. \\[1 em]15\cdot4 \cdot \frac{x}{3} &= 15\cdot2 \cdot \frac{x^2}{5}&& \text{cancel out the denominators} \\[1 em]20x &= 6x^2&& \text{move all terms to the left hand side } \\[1 em]20x-6x^2 &= 0&& \text{simplify left side} \\[1 em]-6x^2+20x &= 0&& \\[1 em] \end{aligned} $$
In order to solve $ \color{blue}{ -6x^{2}+20x = 0 } $, first we need to factor our $ x $.
$$ -6x^{2}+20x = x \left( -6x+20 \right) $$$ x = 0 $ is a root of multiplicity $ 1 $.
The second root can be found by solving equation $ -6x+20 = 0$.
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