$$\frac{1}{x} = 3+\frac{7}{x^2+7x}$$
Answer
$$ \begin{matrix}x_1 = 0 & x_2 = -\dfrac{ 20 }{ 3 } \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} \frac{1}{x} &= 3+\frac{7}{x^2+7x}&& \text{multiply ALL terms by } \color{blue}{ x(x^2+7x) }. \\[1 em]x(x^2+7x)\cdot\frac{1}{x} &= x(x^2+7x)\cdot3+x(x^2+7x)\cdot\frac{7}{x^2+7x}&& \text{cancel out the denominators} \\[1 em]x^2+7x &= 3x^3+21x^2+7x&& \text{move all terms to the left hand side } \\[1 em]x^2+7x-3x^3-21x^2-7x &= 0&& \text{simplify left side} \\[1 em]x^2+7x-3x^3-21x^2-7x &= 0&& \\[1 em]-3x^3-20x^2 &= 0&& \\[1 em] \end{aligned} $$
In order to solve $ \color{blue}{ -3x^{3}-20x^{2} = 0 } $, first we need to factor our $ x^2 $.
$$ -3x^{3}-20x^{2} = x^2 \left( -3x-20 \right) $$$ x = 0 $ is a root of multiplicity $ 2 $.
The second root can be found by solving equation $ -3x-20 = 0$.
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