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Question
$$3x+\frac{9}{x^2}-x-2x^2-5x+\frac{6}{x}+3 = 0$$
Answer
This equation has no solution.
Explanation
$$ \begin{aligned} 3x+\frac{9}{x^2}-x-2x^2-5x+\frac{6}{x}+3 &= 0&& \text{multiply ALL terms by } \color{blue}{ x^2x }. \\[1 em]x^2x\cdot3x+x^2x\cdot\frac{9}{x^2}-x^2xx-x^2x\cdot2x^2-x^2x\cdot5x+x^2x\cdot\frac{6}{x}+x^2x\cdot3 &= x^2x\cdot0&& \text{cancel out the denominators} \\[1 em]3x^4+\frac{9}{x^1}-x^4-2x^5-5x^4+6+3x^3 &= 0&& \text{multiply ALL terms by } \color{blue}{ x^1 }. \\[1 em]x^1\cdot3x^4+x^1\cdot\frac{9}{x^1}-x^1\cdot1x^4-x^1\cdot2x^5-x^1\cdot5x^4+x^1\cdot6+x^1\cdot3x^3 &= x^1\cdot0&& \text{cancel out the denominators} \\[1 em]3x^5+9-x^5-2x^6-5x^5+6x+3x^4 &= 0&& \text{simplify left side} \\[1 em]-2x^6-3x^5+3x^4+6x+9 &= 0&& \\[1 em] \end{aligned} $$
This polynomial has no rational roots that can be found using Rational Root Test.
Roots were found using Newton method.
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