$$5x^2-8x+3(x+2)x = 0$$
Answer
$$ \begin{matrix}x_1 = 0 & x_2 = \dfrac{ 1 }{ 4 } \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} 5x^2-8x+3(x+2)x &= 0&& \text{simplify left side} \\[1 em]5x^2-8x+(3x+6)x &= 0&& \\[1 em]5x^2-8x+3x^2+6x &= 0&& \\[1 em]8x^2-2x &= 0&& \\[1 em] \end{aligned} $$
In order to solve $ \color{blue}{ 8x^{2}-2x = 0 } $, first we need to factor our $ x $.
$$ 8x^{2}-2x = x \left( 8x-2 \right) $$$ x = 0 $ is a root of multiplicity $ 1 $.
The second root can be found by solving equation $ 8x-2 = 0$.
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