Question
$$\frac{1}{4}x+\frac{1}{4}x^2 = x-\frac{3}{2}x^2$$
Answer
$$ \begin{matrix}x_1 = 0 & x_2 = \dfrac{ 3 }{ 7 } \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} \frac{1}{4}x+\frac{1}{4}x^2 &= x-\frac{3}{2}x^2&& \text{multiply ALL terms by } \color{blue}{ 4 }. \\[1 em]4 \cdot \frac{1}{4}x+4\frac{1}{4}x^2 &= 4x-4 \cdot \frac{3}{2}x^2&& \text{cancel out the denominators} \\[1 em]x+x^2 &= 4x-6x^2&& \text{simplify left and right hand side} \\[1 em]x^2+x &= -6x^2+4x&& \text{move all terms to the left hand side } \\[1 em]x^2+x+6x^2-4x &= 0&& \text{simplify left side} \\[1 em]7x^2-3x &= 0&& \\[1 em] \end{aligned} $$
In order to solve $ \color{blue}{ 7x^{2}-3x = 0 } $, first we need to factor our $ x $.
$$ 7x^{2}-3x = x \left( 7x-3 \right) $$$ x = 0 $ is a root of multiplicity $ 1 $.
The second root can be found by solving equation $ 7x-3 = 0$.
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