Question
$$x\cdot2-4 = (x+2)(x-2)$$
Answer
$$ \begin{matrix}x_1 = 0 & x_2 = 2 \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} x\cdot2-4 &= (x+2)(x-2)&& \text{simplify right side} \\[1 em]x\cdot2-4 &= x^2-2x+2x-4&& \\[1 em]x\cdot2-4 &= x^2-2x+2x-4&& \\[1 em]x\cdot2-4 &= x^2-4&& \text{move all terms to the left hand side } \\[1 em]2x-4-x^2+4 &= 0&& \text{simplify left side} \\[1 em]2x-4-x^2+4 &= 0&& \\[1 em]-x^2+2x &= 0&& \\[1 em] \end{aligned} $$
In order to solve $ \color{blue}{ -x^{2}+2x = 0 } $, first we need to factor our $ x $.
$$ -x^{2}+2x = x \left( -x+2 \right) $$$ x = 0 $ is a root of multiplicity $ 1 $.
The second root can be found by solving equation $ -x+2 = 0$.
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