$$\frac{1}{x}(x+1)+\frac{1}{x+1}(x+2) = \frac{1}{12}$$
Answer
$$ \begin{matrix}x_1 = -\dfrac{ 47 }{ 46 }-\dfrac{\sqrt{ 1105 }}{ 46 } & x_2 = -\dfrac{ 47 }{ 46 }+\dfrac{\sqrt{ 1105 }}{ 46 } \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} \frac{1}{x}(x+1)+\frac{1}{x+1}(x+2) &= \frac{1}{12}&& \text{multiply ALL terms by } \color{blue}{ x(x+1)\cdot12 }. \\[1 em]x(x+1)\cdot12 \cdot \frac{1}{x}(x+1)+x(x+1)\cdot12\frac{1}{x+1}(x+2) &= x(x+1)\cdot12\cdot\frac{1}{12}&& \text{cancel out the denominators} \\[1 em]12x^2+24x+12+12x^2+24x &= x^2+x&& \text{simplify left side} \\[1 em]24x^2+48x+12 &= x^2+x&& \text{move all terms to the left hand side } \\[1 em]24x^2+48x+12-x^2-x &= 0&& \text{simplify left side} \\[1 em]23x^2+47x+12 &= 0&& \\[1 em] \end{aligned} $$
$ 23x^{2}+47x+12 = 0 $ is a quadratic equation.
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