check the picture below.
so, we know the dimensions of the pool, is a 20x10, so its area is simply 200 ft², and we know the walkway is 216 ft², so the whole thing, including pool and walkway is really 200 + 216 ft².
now, as you see in the picture, the dimensions for the combined area is 20+2x and 10+2x, since the walkway is "x" long, therefore,
[tex] \bf \stackrel{length}{(20+2x)}\stackrel{width}{(10+2x)}=200+216\implies \stackrel{FOIL}{4x^2+60x+200}=200+216
\\\\\\
4x^2+60x=216\implies \stackrel{dividing~by~4}{x^2+15x=54}
\\\\\\
x^2+15x-54=0\implies (x+18)(x-3)=0\implies x=
\begin{cases}
-18\\
\boxed{3}
\end{cases} [/tex]
notice, it cannot be -18, since is a positive length unit.
