Question
$$2x(x+3)-5x\cdot(6-x) = 8x(3x+3)-4x\cdot(1-2x)$$
Answer
$$ \begin{matrix}x_1 = 0 & x_2 = -\dfrac{ 44 }{ 25 } \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} 2x(x+3)-5x\cdot(6-x) &= 8x(3x+3)-4x\cdot(1-2x)&& \text{simplify left and right hand side} \\[1 em]2x^2+6x-(30x-5x^2) &= 24x^2+24x-(4x-8x^2)&& \\[1 em]2x^2+6x-30x+5x^2 &= 24x^2+24x-4x+8x^2&& \\[1 em]7x^2-24x &= 32x^2+20x&& \text{move all terms to the left hand side } \\[1 em]7x^2-24x-32x^2-20x &= 0&& \text{simplify left side} \\[1 em]-25x^2-44x &= 0&& \\[1 em] \end{aligned} $$
In order to solve $ \color{blue}{ -25x^{2}-44x = 0 } $, first we need to factor our $ x $.
$$ -25x^{2}-44x = x \left( -25x-44 \right) $$$ x = 0 $ is a root of multiplicity $ 1 $.
The second root can be found by solving equation $ -25x-44 = 0$.
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