$$\frac{v+2}{v}+\frac{4}{3v} = 11$$
Answer
$$ \begin{matrix}v_1 = \dfrac{ 15 }{ 4 }-\dfrac{\sqrt{ 201 }}{ 4 } & v_2 = \dfrac{ 15 }{ 4 }+\dfrac{\sqrt{ 201 }}{ 4 } \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} \frac{v+2}{v}+\frac{4}{3v} &= 11&& \text{multiply ALL terms by } \color{blue}{ v\cdot3 }. \\[1 em]v\cdot3 \cdot \frac{v+2}{v}+v\cdot3\cdot\frac{4}{3v} &= v\cdot3\cdot11&& \text{cancel out the denominators} \\[1 em]3v+6+4v^2 &= 33v&& \text{simplify left side} \\[1 em]4v^2+3v+6 &= 33v&& \text{move all terms to the left hand side } \\[1 em]4v^2+3v+6-33v &= 0&& \text{simplify left side} \\[1 em]4v^2-30v+6 &= 0&& \\[1 em] \end{aligned} $$
$ 4x^{2}-30x+6 = 0 $ is a quadratic equation.
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