Question
$$3x-5x(x-7)+4x(3x-4) = 0$$
Answer
$$ \begin{matrix}x_1 = 0 & x_2 = -\dfrac{ 22 }{ 7 } \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} 3x-5x(x-7)+4x(3x-4) &= 0&& \text{simplify left side} \\[1 em]3x-(5x^2-35x)+12x^2-16x &= 0&& \\[1 em]3x-5x^2+35x+12x^2-16x &= 0&& \\[1 em]-5x^2+38x+12x^2-16x &= 0&& \\[1 em]7x^2+22x &= 0&& \\[1 em] \end{aligned} $$
In order to solve $ \color{blue}{ 7x^{2}+22x = 0 } $, first we need to factor our $ x $.
$$ 7x^{2}+22x = x \left( 7x+22 \right) $$$ x = 0 $ is a root of multiplicity $ 1 $.
The second root can be found by solving equation $ 7x+22 = 0$.
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