The parabolas defined by the equations $y=x^2+4x+6$ and $y=\frac{1}{2}x^2+x+6$ intersect at points $(a,b)$ and $(c,d)$, where $c\ge a$. What is $c-a$?
The graph of the two parabolas is shown below:

[asy]
Label f;

f.p=fontsize(4);

xaxis(-7,1,Ticks(f, 2.0));

yaxis(0,25,Ticks(f, 5.0));
real f(real x)

{

return x^2+4x+6;

}

draw(graph(f,-7,1),linewidth(1));
real g(real x)

{

return .5x^2+x+6;

}

draw(graph(g,-7,1),linewidth(1));
[/asy]

The graphs intersect when $y$ equals both $x^2 + 4x +6$ and $\frac12x^2 + x+6$, 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 $c-a=\boxed{6}$.