$$4 \cdot \frac{x}{x}+3 = \frac{12}{x^2}$$
Answer
$$ \begin{matrix}x_1 = 1.20063 & x_2 = -1.20063 & x_3 = 1.66579i \\[1 em] x_4 = -1.66579i & \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} 4 \cdot \frac{x}{x}+3 &= \frac{12}{x^2}&& \text{multiply ALL terms by } \color{blue}{ xx^2 }. \\[1 em]xx^2\cdot4 \cdot \frac{x}{x}+xx^2\cdot3 &= xx^2\cdot\frac{12}{x^2}&& \text{cancel out the denominators} \\[1 em]4x+3x^3 &= \frac{12}{x^1}&& \text{multiply ALL terms by } \color{blue}{ x^1 }. \\[1 em]x^1\cdot4x+x^1\cdot3x^3 &= x^1\cdot\frac{12}{x^1}&& \text{cancel out the denominators} \\[1 em]4x^2+3x^4 &= 12&& \text{simplify left side} \\[1 em]3x^4+4x^2 &= 12&& \text{move all terms to the left hand side } \\[1 em]3x^4+4x^2-12 &= 0&& \\[1 em] \end{aligned} $$
This polynomial has no rational roots that can be found using Rational Root Test.
Roots were found using quartic formulas
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