College Algebra...applying Quadratic Equations...!!! Thank you!?

Area of the pathway in sq. ft:
39*37 - (39 - 2x)(37 - 2x) = 355
39*37 - 39*37 + 152x - 4x² = 355
4x² - 152x + 355 = 0
(2x - 5)(2x - 71) = 0
x = 2.5 or x = 35.5 (rejected)

Width of the pathway = 2.5 ft

We know the total length and width are 37 and 39 feet.

This is a total area of 1443 ft².

We know there are 355 ft² of stones, which results in an area (in green) to be:

1443 - 355 = 1088 ft²

If "x" is the width of the pathway, we need to find expressions for length and width of the green area in terms of x so we can solve for x.

With "x" ft on top and bottom, the length is (37 - 2x) ft. With "x" on the left and right, the width is (39 - 2x) ft.

So now we can multiply these to get the area, which is known, to solve for x:

A = lw
1088 = (37 - 2x)(39 - 2x)
1088 = 1443 - 74x - 78x + 4x²
1088 = 1443 - 152x + 4x²
0 = 355 - 152x + 4x²
0 = 4x² - 152x + 355

We can now use quadratic equation to solve for x:

x = [ -b ± √(b² - 4ac)] / (2a)
x = [ -(-152) ± √((-152)² - 4(4)(355))] / (2 * 4)
x = [ 152 ± √(23104 - 5680)] / 8
x = [ 152 ± √(17424)] / 8
x = (152 ± 132) / 8
x = 20/8 and 284/8
x = 2.5 and 35.5

We obviously can't have 35.5 ft of pathing, since if we double it and subtract it from 37 or 39 we have a negative value.

So that leaves us with 2.5 ft of pathway all the way around.

blue rectangle = 39 * x
blue rectangle = 39x

red rectangle = (37 - x - x) * x
red rectangle = x.(37 - 2x)

You must calculate the surface area of the border, i.e. the grey area.
= (2 * blue rectangle) + (2 * red rectangle)
= 2 * [blue rectangle + red rectangle]
= 2 * [39x + x.(37 - 2x)] → given that it's 355 ft²
2 * [39x + x.(37 - 2x)] = 355
39x + x.(37 - 2x) = 355/2
39x + 37x - 2x² = 355/2
- 2x² + 76x = 355/2
x² - 38x = - 355/4 → completing the square
x² - 38x + 19² = - (355/4) + 19²
x² - 38x + 19² = 1089/4
(x - 19)² = [± (33/2)]²
x - 19 = ± (33/2)
x = 19 ± (33/2)
x = (38 ± 33)/2

First case:
x = (38 + 33)/2
x = 71/2 ← no possible because the blue rectangle doesn't exist

Second case:
x = (38 - 33)/2
x = 5/2 ← ok

With x being the width of the pathway, its total area is 2 * (2x² + x(39 - 2x) + x(37 - 2x)). This is supposed to equal 355 ft². Thus:
4x² + 78x - 4x² + 74x - 4x² = 355
-4x² + 152x - 355 = 0
4x² - 152x + 355= 0

x = (152 ± √((152)² - (4)(4)(355)))/8 = 19 ± 132/8 = 35.5 ft or 2.5 ft. It cannot be 35.5 ft, because that would make the pathway wider and longer than the garden. Therefore the pathway will be 2.5 ft wide.