The P.E of a conservative system is given by  $V\text{}\left(x\right)=(x^2-3x)J$, where x is measured in metre.  Then its equilibrium position is at

$$\begin{gathered} given PE is V\left(x\right)=x^2-3x \\ At equilibrium , the P.E of a system is minimum, \\ to find maximum or minimum value of PE , differentiate PE and equate to zero \\ \therefore \frac{dv}{dx}=0 \\ \Rightarrow \frac{d}{dx}(x^2-3x)=0 \\ \Rightarrow 2x-3=0 \\ \Rightarrow x=\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right. \\ \hspace{0.17em}\hspace{0.17em}\Rightarrow x=1.5m \end{gathered}$$