Question
$$\frac{5}{3x}+\frac{3}{2x^2}-\frac{x-4}{6x^3} = 0$$
Answer
$$ \begin{matrix}x_1 = 7-\sqrt{ 58 } & x_2 = 7+\sqrt{ 58 } \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} \frac{5}{3x}+\frac{3}{2x^2}-\frac{x-4}{6x^3} &= 0&& \text{multiply ALL terms by } \color{blue}{ 3x\cdot2x^2\cdot6x^3 }. \\[1 em]3x\cdot2x^2\cdot6x^3\cdot\frac{5}{3x}+3x\cdot2x^2\cdot6x^3\cdot\frac{3}{2x^2}-3x\cdot2x^2\cdot6x^3\frac{x-4}{6x^3} &= 3x\cdot2x^2\cdot6x^3\cdot0&& \text{cancel out the denominators} \\[1 em]60x+54-(6x^2-24x) &= 0&& \text{simplify left side} \\[1 em]60x+54-6x^2+24x &= 0&& \\[1 em]-6x^2+84x+54 &= 0&& \\[1 em] \end{aligned} $$
$ -6x^{2}+84x+54 = 0 $ is a quadratic equation.
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