When x = 0,the left side is 2 and x = 0 isn't a root of the equation

\(\Rightarrow x\ne0\).Divide both sides by \(x^2\),we have :

\(2x^2-x-5+\dfrac{1}{x}+\dfrac{2}{x^2}=0\).Denote \(t=x-\dfrac{1}{x}\),then we have :

\(\circledast\dfrac{x^2-1}{x}=t\Leftrightarrow x^2-tx-1=0\Leftrightarrow x=\dfrac{t\pm\sqrt{t^2+4}}{2}\)

\(\circledast x^2+\dfrac{1}{x^2}=x^2-2+\dfrac{1}{x^2}+2=\left(x-\dfrac{1}{x}\right)^2+2=t^2+2\)

The equation will change into :

\(2\left(t^2+2\right)-t-5=0\Leftrightarrow2t^2-t-1=0\)

\(\Leftrightarrow\left[{}\begin{matrix}t=\dfrac{1+\sqrt{1+8}}{4}=\dfrac{1+3}{4}=1\\t=\dfrac{1-\sqrt{1+8}}{4}=\dfrac{1-3}{4}=-\dfrac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1\pm\sqrt{1^2+4}}{2}=\dfrac{1\pm\sqrt{5}}{2}\\x=\dfrac{-\dfrac{1}{2}\pm\sqrt{\left(-\dfrac{1}{2}\right)^2+4}}{2}=\dfrac{-1\pm\sqrt{17}}{4}\end{matrix}\right.\)

Hence,\(S=\left\{\dfrac{1\pm\sqrt{5}}{2};\dfrac{-1\pm\sqrt{17}}{4}\right\}\)