Question 514864
Try drawing a picture.  Let x represent the width of the deck.  We can break the deck up into eight pieces: the two rectangles adjacent to the long sides of the pool, the two rectangles adjacent to the short sides of the pool, and the four corners, each of which is a square with sides the width of the deck.  Let's describe the areas of each piece of the deck algebraically:

Each rectangle adjacent to a long side of the pool has length 20 meters and width x meters, and thus area 20x square meters.

Each rectangle adjacent to a short side of the pool has length 10 meters and width x meters, and thus area 10x square meters.

Each square corner has all sides of width x meters, and thus has area x^2 square meters.

Since there are two of each rectangle and four square corners, the total area of the deck is:

{{{ 2(20x) + 2(10x) + 4x^2 }}}

We know that the area of the deck is 216 square meters, so we have a quadratic equation we can solve:

{{{ 2(20x) + 2(10x) + 4x^2 = 216 }}}

Now we solve:

{{{ 40x + 20x + 4x^2 = 216 }}} (simplify the left hand side)
{{{ 60x + 4x^2 = 216 }}} (combine like terms)
{{{ 4x^2 + 60x - 216 = 0 }}} (subtract 216 from both sides and rearrange terms)
{{{ x^2 + 15x - 54 = 0 }}} (divide both sides by 4)
{{{ (x + 18)(x - 3) = 0 }}} (factor the quadratic: 18 and -3 have a sum of 15 and a product of -54)

Setting both factors equal to 0, we get the roots x = -18 and x = 3.  Now x = -18 cannot be the solution, since x represents a width and cannot be negative.  So the only realistic solution is x = 3.  So the deck has a uniform width of 3 meters.