Step 1: Assume a general tangent line.
Let the common tangent be $y=mx+c$. It must touch both parabolas, giving two conditions that pin down $m$ and $c$.
Step 2: Apply tangency to the first parabola.
Substituting into $y=-x^2$ gives $x^2+mx+c=0$. For a tangent the discriminant is zero: $m^2-4c=0$, so $c=\dfrac{m^2}{4}$.
Step 3: Apply tangency to the second parabola.
Substituting into $y=(x-2)^2$ gives $x^2-(m+4)x+(4-c)=0$. Tangency needs $(m+4)^2-4(4-c)=0$.
Step 4: Combine the two conditions.
Put $c=\dfrac{m^2}{4}$ into the second condition: $(m+4)^2-16+m^2=0$. Expanding, $m^2+8m+16-16+m^2=0$, i.e. $2m^2+8m=0$.
Step 5: Solve for $m$.
Factor $2m(m+4)=0$, so $m=0$ or $m=-4$. The slope $m=0$ would give a horizontal line that cannot be tangent to both as required, leaving $m=-4$.
Step 6: Find $c$ and write the line.
Then $c=\dfrac{(-4)^2}{4}=4$, so the common tangent is $y=-4x+4$.
\[ \boxed{y=-4x+4} \]