$$2 \cdot \frac{x}{x+1}+\frac{5}{2x} = 2$$
Answer
$$ \begin{matrix}x_1 = 0 & x_2 = -\dfrac{ 1 }{ 2 }-\dfrac{\sqrt{ 105 }}{ 10 } & x_3 = -\dfrac{ 1 }{ 2 }+\dfrac{\sqrt{ 105 }}{ 10 } \end{matrix} $$
Explanation
$$ \begin{aligned} 2 \cdot \frac{x}{x+1}+\frac{5}{2x} &= 2&& \text{multiply ALL terms by } \color{blue}{ (x+1)\cdot2x }. \\[1 em](x+1)\cdot2x\cdot2 \cdot \frac{x}{x+1}+(x+1)\cdot2x\cdot\frac{5}{2x} &= (x+1)\cdot2x\cdot2&& \text{cancel out the denominators} \\[1 em]4x^2+5x^3+5x^2 &= 4x^2+4x&& \text{simplify left side} \\[1 em]5x^3+9x^2 &= 4x^2+4x&& \text{move all terms to the left hand side } \\[1 em]5x^3+9x^2-4x^2-4x &= 0&& \text{simplify left side} \\[1 em]5x^3+5x^2-4x &= 0&& \\[1 em] \end{aligned} $$
In order to solve $ \color{blue}{ 5x^{3}+5x^{2}-4x = 0 } $, first we need to factor our $ x $.
$$ 5x^{3}+5x^{2}-4x = x \left( 5x^{2}+5x-4 \right) $$$ x = 0 $ is a root of multiplicity $ 1 $.
The remaining roots can be found by solving equation $ 5x^{2}+5x-4 = 0$.
$ 5x^{2}+5x-4 = 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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