Given:
The area of the playground is 204 square yd.
The width of the playground is 5 yd longer than its length.
Let, w be the with of the playground and l is length.
[tex]w=5+l[/tex]
The area is given as,
[tex]\begin{gathered} A=l\times w \\ 204=l\times(l+5) \\ l^2+5l-204=0 \\ Use\text{ quadratic formula:} \\ l=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a},a=1,b=5,c=-204 \\ l=\frac{-5\pm\sqrt[]{5^2-4\times1\times(-204)}}{2} \\ l=\frac{-5\pm\: 29}{2} \\ l=\frac{-5+29}{2},l=\frac{-5-29}{2} \\ l=12,l=-17 \end{gathered}[/tex]
As length can not be negative.
Hence, length = l =12 yd
Width = w = l +5 = 12+5 =17 yd.
Answer: owidth is 17 yds.
