$$2 \cdot \frac{x}{x-3}+\frac{1}{2x+3}+\frac{3x+9}{(x-3)(2x+3)} = 1$$
Answer
$$ \begin{matrix}x_1 = -1.14401 & x_2 = -2.51133+0.17493i & x_3 = -2.51133-0.17493i \end{matrix} $$
Explanation
$$ \begin{aligned} 2 \cdot \frac{x}{x-3}+\frac{1}{2x+3}+\frac{3x+9}{(x-3)(2x+3)} &= 1&& \text{multiply ALL terms by } \color{blue}{ (x-3)(2x+3) }. \\[1 em](x-3)(2x+3)\cdot2 \cdot \frac{x}{x-3}+(x-3)(2x+3)\cdot\frac{1}{2x+3}+(x-3)(2x+3)\frac{3x+9}{(x-3)(2x+3)} &= (x-3)(2x+3)\cdot1&& \text{cancel out the denominators} \\[1 em]4x^2+6x+x-3+12x^3+72x^2+135x+81 &= 2x^2-3x-9&& \text{simplify left side} \\[1 em]4x^2+7x-3+12x^3+72x^2+135x+81 &= 2x^2-3x-9&& \\[1 em]12x^3+76x^2+142x+78 &= 2x^2-3x-9&& \text{move all terms to the left hand side } \\[1 em]12x^3+76x^2+142x+78-2x^2+3x+9 &= 0&& \text{simplify left side} \\[1 em]12x^3+74x^2+145x+87 &= 0&& \\[1 em] \end{aligned} $$
This polynomial has no rational roots that can be found using Rational Root Test.
Roots were found using qubic formulas.
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