$$\frac{1}{6}x^2 = \frac{1}{2}x+\frac{7}{6}x^2$$
Answer
$$ \begin{matrix}x_1 = 0 & x_2 = -\dfrac{ 1 }{ 2 } \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} \frac{1}{6}x^2 &= \frac{1}{2}x+\frac{7}{6}x^2&& \text{multiply ALL terms by } \color{blue}{ 6 }. \\[1 em]6 \cdot \frac{1}{6}x^2 &= 6 \cdot \frac{1}{2}x+6\frac{7}{6}x^2&& \text{cancel out the denominators} \\[1 em]x^2 &= 3x+7x^2&& \text{simplify right side} \\[1 em]x^2 &= 7x^2+3x&& \text{move all terms to the left hand side } \\[1 em]x^2-7x^2-3x &= 0&& \text{simplify left side} \\[1 em]-6x^2-3x &= 0&& \\[1 em] \end{aligned} $$
In order to solve $ \color{blue}{ -6x^{2}-3x = 0 } $, first we need to factor our $ x $.
$$ -6x^{2}-3x = x \left( -6x-3 \right) $$$ x = 0 $ is a root of multiplicity $ 1 $.
The second root can be found by solving equation $ -6x-3 = 0$.
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