$$\frac{300}{x^7}-300x^{12}-2 = 0$$
Answer
$$ \begin{matrix}x_1 = 0.99965 & x_2 = 0.54698+0.83682i & x_3 = 0.54698-0.83682i \\[1 em] x_4 = -0.67761+0.73584i & x_5 = -0.67761-0.73584i & x_6 = -0.40135+0.91583i \\[1 em] x_7 = -0.40135-0.91583i & x_8 = 0.78895+0.61451i & x_9 = 0.78895-0.61451i \\[1 em] x_10 = 0.24563+0.96972i & x_11 = 0.24563-0.96972i & x_12 = -0.87924+0.47569i \\[1 em] x_13 = -0.87924-0.47569i & x_14 = -0.08286+0.99637i & x_15 = -0.08286-0.99637i \\[1 em] x_16 = 0.94613+0.32453i & x_17 = 0.94613-0.32453i & x_18 = -0.98645+0.16493i \\[1 em] x_19 = -0.98645-0.16493i & \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} \frac{300}{x^7}-300x^{12}-2 &= 0&& \text{multiply ALL terms by } \color{blue}{ x^7 }. \\[1 em]x^7\cdot\frac{300}{x^7}-x^7\cdot300x^{12}-x^7\cdot2 &= x^7\cdot0&& \text{cancel out the denominators} \\[1 em]300-300x^{19}-2x^7 &= 0&& \text{simplify left side} \\[1 em]-300x^{19}-2x^7+300 &= 0&& \\[1 em] \end{aligned} $$
This polynomial has no rational roots that can be found using Rational Root Test.
Roots were found using Newton method.
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