Question
$$\frac{44}{7}\cdot2x\cdot2\cdot(29-2x)+2\frac{22}{7}x^2 = 0$$
Answer
$$ \begin{matrix}x_1 = 0 & x_2 = \dfrac{ 116 }{ 7 } \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} \frac{44}{7}\cdot2x\cdot2\cdot(29-2x)+2\frac{22}{7}x^2 &= 0&& \text{multiply ALL terms by } \color{blue}{ 7 }. \\[1 em]7 \cdot \frac{44}{7}\cdot2x\cdot2\cdot(29-2x)+7\cdot2\frac{22}{7}x^2 &= 7\cdot0&& \text{cancel out the denominators} \\[1 em]-352x^2+5104x+44x^2 &= 0&& \text{simplify left side} \\[1 em]-308x^2+5104x &= 0&& \\[1 em] \end{aligned} $$
In order to solve $ \color{blue}{ -308x^{2}+5104x = 0 } $, first we need to factor our $ x $.
$$ -308x^{2}+5104x = x \left( -308x+5104 \right) $$$ x = 0 $ is a root of multiplicity $ 1 $.
The second root can be found by solving equation $ -308x+5104 = 0$.
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