$$x^3-4x^2+x+\frac{9}{x}-2 = 0$$
Answer
$$ \begin{matrix}x_1 = 1.48494 & x_2 = 3.69784 & x_3 = -0.59139+1.13547i \\[1 em] x_4 = -0.59139-1.13547i & \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} x^3-4x^2+x+\frac{9}{x}-2 &= 0&& \text{multiply ALL terms by } \color{blue}{ x }. \\[1 em]xx^3-x\cdot4x^2+xx+x\cdot\frac{9}{x}-x\cdot2 &= x\cdot0&& \text{cancel out the denominators} \\[1 em]x^4-4x^3+x^2+9-2x &= 0&& \text{simplify left side} \\[1 em]x^4-4x^3+x^2-2x+9 &= 0&& \\[1 em] \end{aligned} $$
This polynomial has no rational roots that can be found using Rational Root Test.
Roots were found using quartic formulas
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