$$\frac{7x+1}{2x+5}+1 = \frac{10x-3}{3x}$$
Answer
$$ \begin{matrix}x_1 = 0 & x_2 = -0.33104 & x_3 = 0.96077 \\[1 em] x_4 = -2.82972 & \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} \frac{7x+1}{2x+5}+1 &= \frac{10x-3}{3x}&& \text{multiply ALL terms by } \color{blue}{ (2x+5)\cdot3x }. \\[1 em](2x+5)\cdot3x \cdot \frac{7x+1}{2x+5}+(2x+5)\cdot3x\cdot1 &= (2x+5)\cdot3x \cdot \frac{10x-3}{3x}&& \text{cancel out the denominators} \\[1 em]21x^2+3x+6x^2+15x &= 20x^4+44x^3-15x^2&& \text{simplify left side} \\[1 em]27x^2+18x &= 20x^4+44x^3-15x^2&& \text{move all terms to the left hand side } \\[1 em]27x^2+18x-20x^4-44x^3+15x^2 &= 0&& \text{simplify left side} \\[1 em]-20x^4-44x^3+42x^2+18x &= 0&& \\[1 em] \end{aligned} $$
In order to solve $ \color{blue}{ -20x^{4}-44x^{3}+42x^{2}+18x = 0 } $, first we need to factor our $ x $.
$$ -20x^{4}-44x^{3}+42x^{2}+18x = x \left( -20x^{3}-44x^{2}+42x+18 \right) $$$ x = 0 $ is a root of multiplicity $ 1 $.
The remaining roots can be found by solving equation $ -20x^{3}-44x^{2}+42x+18 = 0$.
This polynomial has no rational roots that can be found using Rational Root Test.
Roots were found using qubic formulas.
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