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# Advanced Math/Lenght, Width - Solution / Explanation

Question
A rectangular park is enclosed a walkway 4 yards wide.

Park- 15200 square yards.

Walkway- 2104 square yards.

Determine length and width of park.

Let x = length of park.

15200 / x = width.

(x + 4)(15200 / x + 8) = (15200 + 2104)= 17304 **

8x + 121600 / x + 15264 = 17304 **

8x^2 - 2040x = 121600 **

x^2 - 255x = 15200

x = 127 * 5 + - 32 * 5 = 160 or 95

length of park = 160 yards or 95 yards.

width of park = 15200 / 160 = 95 or 15200 / 95 = 160

** I don't understand these steps.

Also, can you provide some hints/tips on quadratics?  I am not very good at these.  Thanks.

Park and walkway
There is an error in the solution.
The length of the park is x.
The width of the park is 15200/x.

Since the walkway is 4 yd wide, it adds a total of 8 yds to both length and width. https://www.flickr.com/photos/dwread/15884823021/

Combined, the area of the park plus walkway is 15200+2104 = 17304 yd².
Therefore, the equation for the combined area is
(x+8)((15,200/x)+8) = 17,304

expand the equation
15200 + 8·15200/x + 8x + 8² = 17304

factor 8 out of each term
1900 + 15200/x + x + 8 = 2163
-255 + 15200/x + x = 0

multiply both sides by x
-255x + 15200 + x² = 0

rearrange to standard form
x² - 255x + 15200 = 0

x = [255 ± √(255² – 4·1·15200)] / [2·1]
= [255 ± √4225] / 2
= [255 ± 65] /2
= 95, 160

The dimensions of the park are 95 yards by 160 yards.

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