Question
$$\frac{7}{2}x+\frac{141}{2} = \frac{7}{x}-10$$
Answer
$$ \begin{matrix}x_1 = -\dfrac{ 23 }{ 2 }-\dfrac{\sqrt{ 537 }}{ 2 } & x_2 = -\dfrac{ 23 }{ 2 }+\dfrac{\sqrt{ 537 }}{ 2 } \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} \frac{7}{2}x+\frac{141}{2} &= \frac{7}{x}-10&& \text{multiply ALL terms by } \color{blue}{ 2x }. \\[1 em]2x \cdot \frac{7}{2}x+2x\cdot\frac{141}{2} &= 2x\cdot\frac{7}{x}-2x\cdot10&& \text{cancel out the denominators} \\[1 em]7x^2+141x &= 14-20x&& \text{move all terms to the left hand side } \\[1 em]7x^2+141x-14+20x &= 0&& \text{simplify left side} \\[1 em]7x^2+161x-14 &= 0&& \\[1 em] \end{aligned} $$
$ 7x^{2}+161x-14 = 0 $ is a quadratic equation.
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