Question
$$1-\frac{\frac{3}{x}}{x+2}-\frac{x^2}{x}+1 = 0$$
Answer
$$ \begin{matrix}x_1 = 1 & x_2 = -\dfrac{ 5 }{ 2 } \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} 1-\frac{\frac{3}{x}}{x+2}-\frac{x^2}{x}+1 &= 0&& \text{multiply ALL terms by } \color{blue}{ (x+2)x }. \\[1 em](x+2)x\cdot1-(x+2)x \cdot \frac{\frac{3}{x}}{x+2}-(x+2)x\frac{x^2}{x}+(x+2)x\cdot1 &= (x+2)x\cdot0&& \text{cancel out the denominators} \\[1 em]x^2+2x-3-(x+2)+x^2+2x &= 0&& \text{simplify left side} \\[1 em]x^2+2x-3-x-2+x^2+2x &= 0&& \\[1 em]x^2+x-5+x^2+2x &= 0&& \\[1 em]2x^2+3x-5 &= 0&& \\[1 em] \end{aligned} $$
$ 2x^{2}+3x-5 = 0 $ is a quadratic equation.
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