$$-\frac{1}{6}x^2+\frac{1}{4}x = \frac{1}{x}$$
Answer
$$ \begin{matrix}x_1 = \dfrac{ 1 }{ 3 }-\dfrac{\sqrt{ 37 }}{ 3 } & x_2 = \dfrac{ 1 }{ 3 }+\dfrac{\sqrt{ 37 }}{ 3 } \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} -\frac{1}{6}x^2+\frac{1}{4}x &= \frac{1}{x}&& \text{multiply ALL terms by } \color{blue}{ 6\cdot4x }. \\[1 em]-6\cdot4x \cdot \frac{1}{6}x^2+6\cdot4x\frac{1}{4}x &= 6\cdot4x\cdot\frac{1}{x}&& \text{cancel out the denominators} \\[1 em]-4x+6x^2 &= 24&& \text{simplify left side} \\[1 em]6x^2-4x &= 24&& \text{move all terms to the left hand side } \\[1 em]6x^2-4x-24 &= 0&& \\[1 em] \end{aligned} $$
$ 6x^{2}-4x-24 = 0 $ is a quadratic equation.
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