$$\frac{x}{x}+3+\frac{2}{x}-2 = \frac{x^2}{x^2}-x-6$$
Answer
$$ \begin{matrix}x_1 = 0.36644 & x_2 = -0.74585 & x_3 = -7.01412 \\[1 em] x_4 = 0.19676+0.69493i & x_5 = 0.19676-0.69493i \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} \frac{x}{x}+3+\frac{2}{x}-2 &= \frac{x^2}{x^2}-x-6&& \text{multiply ALL terms by } \color{blue}{ xx^2 }. \\[1 em]xx^2\frac{x}{x}+xx^2\cdot3+xx^2\cdot\frac{2}{x}-xx^2\cdot2 &= xx^2\frac{x^2}{x^2}-xx^2x-xx^2\cdot6&& \text{cancel out the denominators} \\[1 em]x+3x^3+2-2x^3 &= \frac{1}{x^1}-x^4-6x^3&& \text{multiply ALL terms by } \color{blue}{ x^1 }. \\[1 em]x^1\cdot1x+x^1\cdot3x^3+x^1\cdot2-x^1\cdot2x^3 &= x^1\cdot\frac{1}{x^1}-x^1\cdot1x^4-x^1\cdot6x^3&& \text{cancel out the denominators} \\[1 em]x^2+3x^4+2x-2x^4 &= 1-x^5-6x^4&& \text{simplify left and right hand side} \\[1 em]x^4+x^2+2x &= -x^5-6x^4+1&& \text{move all terms to the left hand side } \\[1 em]x^4+x^2+2x+x^5+6x^4-1 &= 0&& \text{simplify left side} \\[1 em]x^5+7x^4+x^2+2x-1 &= 0&& \\[1 em] \end{aligned} $$
This polynomial has no rational roots that can be found using Rational Root Test.
Roots were found using Newton method.
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