Question
$$\frac{3}{6}x-5\frac{x}{10}x^3 = \frac{4}{x^2}$$
Answer
$$ \begin{matrix}x_1 = \dfrac{ 1 }{ 2 }+\dfrac{\sqrt{ 31 }}{ 2 }i & x_2 = \dfrac{ 1 }{ 2 }-\dfrac{\sqrt{ 31 }}{ 2 }i \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} \frac{3}{6}x-5\frac{x}{10}x^3 &= \frac{4}{x^2}&& \text{multiply ALL terms by } \color{blue}{ 6\cdot10x^2 }. \\[1 em]6\cdot10x^2\frac{3}{6}x-6\cdot10x^25\frac{x}{10}x^3 &= 6\cdot10x^2\cdot\frac{4}{x^2}&& \text{cancel out the denominators} \\[1 em]30x-30x^2 &= 240&& \text{simplify left side} \\[1 em]-30x^2+30x &= 240&& \text{move all terms to the left hand side } \\[1 em]-30x^2+30x-240 &= 0&& \\[1 em] \end{aligned} $$
$ -30x^{2}+30x-240 = 0 $ is a quadratic equation.
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