Question
$$\frac{1}{3}x^2 = x+\frac{3}{2}x^2-\frac{1}{6}x^2$$
Answer
$$ \begin{matrix}x_1 = 0 & x_2 = -1 \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} \frac{1}{3}x^2 &= x+\frac{3}{2}x^2-\frac{1}{6}x^2&& \text{multiply ALL terms by } \color{blue}{ 6 }. \\[1 em]6 \cdot \frac{1}{3}x^2 &= 6x+6 \cdot \frac{3}{2}x^2-6\frac{1}{6}x^2&& \text{cancel out the denominators} \\[1 em]2x^2 &= 6x+9x^2-x^2&& \text{simplify right side} \\[1 em]2x^2 &= 8x^2+6x&& \text{move all terms to the left hand side } \\[1 em]2x^2-8x^2-6x &= 0&& \text{simplify left side} \\[1 em]-6x^2-6x &= 0&& \\[1 em] \end{aligned} $$
In order to solve $ \color{blue}{ -6x^{2}-6x = 0 } $, first we need to factor our $ x $.
$$ -6x^{2}-6x = x \left( -6x-6 \right) $$$ x = 0 $ is a root of multiplicity $ 1 $.
The second root can be found by solving equation $ -6x-6 = 0$.
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