The length of a rectangular garden is 8 feet longer than the width. the garden is surrounded by a 4-foot sidewalk. the sidewalk has an area of 320 square feet. find the dimensions of the garden.

1. The information given in the problem is:

 - The length of a rectangular garden is 8 feet longer than the width.
 - The garden is surrounded by a 4-foot sidewalk.
 - The area of the sidewalk is 320 ft².

 2. So, the length of the rectangular garden is:

 L1=8+W1

 3. The formula for calculate the area of the sidewalk, is:

 A2=L2xW2

 "A2" is the area of the sidewalk (A2=320 ft²).

 "L2" is the length of the sidewalk.

 "W2" is the widht of the sidewalk.

 4. The length of the sidewalk (L2) is:

 L2=L1+4+4    (4 feet on each side)
 L2=L1+8

 5. When you substitute L1=8+W1 into the equation L2=L1+8, you obtain:

 L2=8+W1+8
 L2=W1+16

 6. The widht of the sidewalk is:

 W2=W1+4+4
 W2=W1+8

 7. Now, you must substitute the length and the widht of the sidewalk into the formula A2=L2xW2:

 A2=L2xW2
 A2=(W1+16)(W1+8)
 320=W1²+16W1+8W1+128
 W1²+16W1+8W1+128-320=0
 W1²+16W1+8W1-192=0

 8. When you solve the quadratic equation, you obtain the value of W1:

 W1=16.97 ft

 9. Finally, you must substitute the value of W1 into the formula L1=8+W1:

 L1=8+W1
 L1=8+16.97 
 L1=24.97 ft

 10. Therefore, the dimensions of the garden are:

 L1=24.97 ft
 W1=16.97 ft

Im sorry i cant answer it : ( im so sorry