We need to solve the given equation and then find the sum of all possible solutions.
The equation is:
[tex]\frac{18-3w}{w+6}=\frac{w^2}{w+6}[/tex]Notice that the denominator on both sides is w+6. Since the denominator can't be zero, we have:
[tex]\begin{gathered} w+6\ne0 \\ \\ w\ne-6 \end{gathered}[/tex]Thus, -6 can't be a solution.
Now, we can solve the equation by rewriting it as
[tex]\begin{gathered} \frac{18-3w}{w+6}\cdot(w+6)=\frac{w^2}{w+6}\cdot(w+6) \\ \\ 18-3w=w^{2} \\ \\ w^{2}+3w-18=0 \end{gathered}[/tex]Now, we can use the quadratic formula to solve it:
[tex]\begin{gathered} w=\frac{-3\pm\sqrt[]{3^{2}-4(1)(-18)}}{2(1)} \\ \\ w=\frac{-3\pm\sqrt[]{9+72}}{2} \\ \\ w=\frac{-3\pm\sqrt[]{81}}{2} \\ \\ w=\frac{-3\pm9}{2} \\ \\ w_1=\frac{-3-9}{2}=-6\text{ (this solution is not possible)} \\ \\ w_2=\frac{-3+9}{2}=3 \end{gathered}[/tex]Therefore, the only possible solution is 3. And the sum of all possible solutions is 3.