$$\frac{45}{x^2}-\frac{4}{x}-1 = 0$$
Answer
$$ \begin{matrix}x_1 = 2.43669 & x_2 = -2.73482 & x_3 = 0.14907+2.59433i \\[1 em] x_4 = 0.14907-2.59433i & \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} \frac{45}{x^2}-\frac{4}{x}-1 &= 0&& \text{multiply ALL terms by } \color{blue}{ x^2x }. \\[1 em]x^2x\cdot\frac{45}{x^2}-x^2x\cdot\frac{4}{x}-x^2x\cdot1 &= x^2x\cdot0&& \text{cancel out the denominators} \\[1 em]\frac{45}{x^1}-4-x^3 &= 0&& \text{multiply ALL terms by } \color{blue}{ x^1 }. \\[1 em]x^1\cdot\frac{45}{x^1}-x^1\cdot4-x^1\cdot1x^3 &= x^1\cdot0&& \text{cancel out the denominators} \\[1 em]45-4x-x^4 &= 0&& \text{simplify left side} \\[1 em]-x^4-4x+45 &= 0&& \\[1 em] \end{aligned} $$
This polynomial has no rational roots that can be found using Rational Root Test.
Roots were found using quartic formulas
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