$$\frac{x^2+3}{7x} = \frac{x+1}{6}$$
Answer
$$ \begin{matrix}x_1 = 0 & x_2 = 0.54706 & x_3 = -0.27353+1.4345i \\[1 em] x_4 = -0.27353-1.4345i & \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} \frac{x^2+3}{7x} &= \frac{x+1}{6}&& \text{multiply ALL terms by } \color{blue}{ 7x\cdot6 }. \\[1 em]7x\cdot6 \cdot \frac{x^2+3}{7x} &= 7x\cdot6 \cdot \frac{x+1}{6}&& \text{cancel out the denominators} \\[1 em]6x^4+18x^2 &= 7x^2+7x&& \text{move all terms to the left hand side } \\[1 em]6x^4+18x^2-7x^2-7x &= 0&& \text{simplify left side} \\[1 em]6x^4+11x^2-7x &= 0&& \\[1 em] \end{aligned} $$
In order to solve $ \color{blue}{ 6x^{4}+11x^{2}-7x = 0 } $, first we need to factor our $ x $.
$$ 6x^{4}+11x^{2}-7x = x \left( 6x^{3}+11x-7 \right) $$$ x = 0 $ is a root of multiplicity $ 1 $.
The remaining roots can be found by solving equation $ 6x^{3}+11x-7 = 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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