$$(3x+2)(x-6)+(4+2x)\cdot(3-x) = 0$$
Answer
$$ \begin{matrix}x_1 = 0 & x_2 = 14 \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} (3x+2)(x-6)+(4+2x)\cdot(3-x) &= 0&& \text{simplify left side} \\[1 em]3x^2-18x+2x-12+12-4x+6x-2x^2 &= 0&& \\[1 em]3x^2-16x-12-2x^2+2x+12 &= 0&& \\[1 em]x^2-14x &= 0&& \\[1 em] \end{aligned} $$
In order to solve $ \color{blue}{ x^{2}-14x = 0 } $, first we need to factor our $ x $.
$$ x^{2}-14x = x \left( x-14 \right) $$$ x = 0 $ is a root of multiplicity $ 1 $.
The second root can be found by solving equation $ x-14 = 0$.
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