$6{{x}^{2}}+13x+4=\frac{3}{x}\Rightarrow 6{{x}^{3}}+13{{x}^{2}}+4x-3=0$

چون $f(-1)=0$ است، پس $f(x)$ یک عامل $(x+1)$ دارد. با انجام تقسیم، عامل دیگر را پیدا می‌کنیم.

$\begin{align}
  & f(x)=(x+1)(6{{x}^{2}}+7x-3)=0 \\
 & \left\{ \begin{matrix}
   x+1=0\Rightarrow {{x}_{3}}=-1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,  \\
   6{{x}^{2}}+7x-3=0\Rightarrow x_{1}^{2}+x_{2}^{2}={{S}^{2}}-2P={{(-\frac{7}{6})}^{2}}-2(-\frac{1}{2})=\frac{85}{36}  \\
\end{matrix} \right. \\
\end{align}$

مجموع مربعاتات ریشه‌ها: $A={{(-1)}^{2}}+\frac{85}{36}=\frac{121}{36}\Rightarrow \sqrt{A}=\frac{11}{6}$