$$x\cdot\frac{1}{2}+2x^2 = 0$$
Answer
$$ \begin{matrix}x_1 = 0 & x_2 = -\dfrac{ 1 }{ 4 } \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} x\cdot\frac{1}{2}+2x^2 &= 0&& \text{multiply ALL terms by } \color{blue}{ 2 }. \\[1 em]2x\cdot\frac{1}{2}+2\cdot2x^2 &= 2\cdot0&& \text{cancel out the denominators} \\[1 em]x+4x^2 &= 0&& \text{simplify left side} \\[1 em]4x^2+x &= 0&& \\[1 em] \end{aligned} $$
In order to solve $ \color{blue}{ 4x^{2}+x = 0 } $, first we need to factor our $ x $.
$$ 4x^{2}+x = x \left( 4x+1 \right) $$$ x = 0 $ is a root of multiplicity $ 1 $.
The second root can be found by solving equation $ 4x+1 = 0$.
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