Question
$$\frac{(5+3)^2}{2x\cdot4}-2 = 0$$
Answer
$$ \begin{matrix}x_1 = 0 & x_2 = \dfrac{ 1 }{ 64 } \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} \frac{(5+3)^2}{2x\cdot4}-2 &= 0&& \text{multiply ALL terms by } \color{blue}{ 2x\cdot4 }. \\[1 em]2x\cdot4 \cdot \frac{(5+3)^2}{2x\cdot4}-2x\cdot4\cdot2 &= 2x\cdot4\cdot0&& \text{cancel out the denominators} \\[1 em]1024x^2-16x &= 0&& \\[1 em] \end{aligned} $$
In order to solve $ \color{blue}{ 1024x^{2}-16x = 0 } $, first we need to factor our $ x $.
$$ 1024x^{2}-16x = x \left( 1024x-16 \right) $$$ x = 0 $ is a root of multiplicity $ 1 $.
The second root can be found by solving equation $ 1024x-16 = 0$.
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