jmc

We note that $(2x{)}^2+2\cdot 37\cdot 2x+37^2=(2x+37{)}^2$. In order for this expression to be a multiple of 47, $2x+37$ must be a multiple of 47. Since we want the least positive value of $x$, we will want $2x+37=47$. It follows that $x=\boxed{5}$.