jmc

The graph of the two parabolas is shown below:



The graphs intersect when $y$ equals both $x^2+4x+6$ and $\frac{1}{2}{x}^2+x+6$, so we have $x^2+4x+6=\frac{1}{2}{x}^2+x+6$. Combining like terms, we get $\frac{1}{2}{x}^2+3x=0$. Factoring out a $x$, we have $x(\frac{1}{2}x+3)=0$. So either $x=0$ or $\frac{1}{2}x+3=0\Rightarrow x=-6$, which are the two $x$ coordinates of the points of intersection. Thus, $c=0$ and $a=-6$, and $c-a=\boxed{6}$.