#### Answer

$x=2.42$ $meters$
This also indicates that at least 2-meter-wide border around the pool can be done, in accordance with the zoning law, using the available tile.

#### Work Step by Step

The shaded region is the tile border with uniform width of $x$.
Set up the equation.
$A_{pool} =$ $LW$
$A_{pool} =$ $(12)(8)$
$A_{pool} =96$
$A_{tile border} =L_{tb}W_{tb}-A_{pool}$
$120 =(12+x)(8+x)-96$
$120 =96+24x+16x+4x^2-96$
$120 =40x+4x^2$
Divide both sides by $4$. The equation becomes:
$30 =10x+x^2$
$x^2+10x-30=0$
Use the quadratic formula: $x = \frac{-b±\sqrt{b^2-4ac}}{2a}$
$a=1$, $b=10$, $c=-30$
$x = \frac{-10±\sqrt{10^2-(4⋅1⋅-30)}}{2⋅1}$
$x = \frac{-10±\sqrt{100-(-120)}}{2}$
$x = \frac{-10±\sqrt{100+120}}{2}$
$x = \frac{-10±\sqrt{220}}{2}$
$x = \frac{-10±\sqrt{4⋅55}}{2}$
$x = \frac{-10±2\sqrt{55}}{2}$
$x=-5±\sqrt {55}$
Since measurements cannot be negative, therefore:
$x=-5+\sqrt {55}$
$x=2.42$ $meters$
This also indicates that at least 2-meter-wide border around the pool can be done, in accordance with the zoning law, using the available tile.