$$\frac{2x+1}{2x-3}+2 = \frac{1}{2x}$$
Answer
$$ \begin{matrix}x_1 = 0 & x_2 = \dfrac{ 15 }{ 4 }-\dfrac{\sqrt{ 145 }}{ 4 } & x_3 = \dfrac{ 15 }{ 4 }+\dfrac{\sqrt{ 145 }}{ 4 } \end{matrix} $$
Explanation
$$ \begin{aligned} \frac{2x+1}{2x-3}+2 &= \frac{1}{2x}&& \text{multiply ALL terms by } \color{blue}{ (2x-3)\cdot2x }. \\[1 em](2x-3)\cdot2x \cdot \frac{2x+1}{2x-3}+(2x-3)\cdot2x\cdot2 &= (2x-3)\cdot2x\cdot\frac{1}{2x}&& \text{cancel out the denominators} \\[1 em]4x^2+2x+8x^2-12x &= 2x^3-3x^2&& \text{simplify left side} \\[1 em]12x^2-10x &= 2x^3-3x^2&& \text{move all terms to the left hand side } \\[1 em]12x^2-10x-2x^3+3x^2 &= 0&& \text{simplify left side} \\[1 em]-2x^3+15x^2-10x &= 0&& \\[1 em] \end{aligned} $$
In order to solve $ \color{blue}{ -2x^{3}+15x^{2}-10x = 0 } $, first we need to factor our $ x $.
$$ -2x^{3}+15x^{2}-10x = x \left( -2x^{2}+15x-10 \right) $$$ x = 0 $ is a root of multiplicity $ 1 $.
The remaining roots can be found by solving equation $ -2x^{2}+15x-10 = 0$.
$ -2x^{2}+15x-10 = 0 $ is a quadratic equation.
You can use step-by-step quadratic equation solver to see a detailed explanation on how to solve this equation.
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