$$\frac{5x+20}{2x^2}+\frac{x-5}{2x^2} = \frac{2}{x}$$
Answer
$$ \begin{matrix}x_1 = -\dfrac{ 5 }{ 4 }-\dfrac{\sqrt{ 321 }}{ 12 } & x_2 = -\dfrac{ 5 }{ 4 }+\dfrac{\sqrt{ 321 }}{ 12 } \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} \frac{5x+20}{2x^2}+\frac{x-5}{2x^2} &= \frac{2}{x}&& \text{multiply ALL terms by } \color{blue}{ 2x^2x }. \\[1 em]2x^2x \cdot \frac{5x+20}{2x^2}+2x^2x\frac{x-5}{2x^2} &= 2x^2x\cdot\frac{2}{x}&& \text{cancel out the denominators} \\[1 em]5x^2+20x+x^2-5x &= 4&& \text{simplify left side} \\[1 em]6x^2+15x &= 4&& \text{move all terms to the left hand side } \\[1 em]6x^2+15x-4 &= 0&& \\[1 em] \end{aligned} $$
$ 6x^{2}+15x-4 = 0 $ is a quadratic equation.
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