$$(2x+5)(x-4) = (x-5)\cdot(4-x)$$
Answer
$$ \begin{matrix}x_1 = 0 & x_2 = 4 \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} (2x+5)(x-4) &= (x-5)\cdot(4-x)&& \text{simplify left and right hand side} \\[1 em]2x^2-8x+5x-20 &= 4x-x^2-20+5x&& \\[1 em]2x^2-3x-20 &= -x^2+9x-20&& \text{move all terms to the left hand side } \\[1 em]2x^2-3x-20+x^2-9x+20 &= 0&& \text{simplify left side} \\[1 em]2x^2-3x-20+x^2-9x+20 &= 0&& \\[1 em]3x^2-12x &= 0&& \\[1 em] \end{aligned} $$
In order to solve $ \color{blue}{ 3x^{2}-12x = 0 } $, first we need to factor our $ x $.
$$ 3x^{2}-12x = x \left( 3x-12 \right) $$$ x = 0 $ is a root of multiplicity $ 1 $.
The second root can be found by solving equation $ 3x-12 = 0$.
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