Help put these word problems in equation

1) The length of a photograph is 3cm less than twice the width. The area is 54cm^2. Find the dimensions of the photograph.

2) The dimensions of a rectangular garden were 5m. Each dimension was increased by the same amount. The garden then had an area of 120m^2. Find the dimensions of the new garden.

3) A rectangular pool measures 5yd by deck of uniform width is constructed around the pool. The deck and pool together cover an area of 72yd^2. How wide is the deck?

Thanks for your help :D

1)
L = W - 3
L*W = 54
therefore
W(W-3) = 54
W^2 - 3W - 54 = 0
(W+6)(W-9) = 0
W = -6 or 9 obviously it can't be -6, so
W = 9 cm
L = W - 3 = 6 cm

2)
if it is rectangular, you need the dimension of the other side
if it square then
(5+x)^2 = 120
x^2 + 10x + 25 = 120
x^2 + 10x - 95 = 0
x = 5.95 or -15.95 obviously the positive answer is the one to use
so the new dimensions of the square plot are
5+5.95 = 10.95 m on each side

3)
not enough information - you've left something out

1) Okay, so let x = width of the photograph.

(2x-3)*x = 54cm^2
(length * width = area)

2) Sounds like it's actually a square, since all dimensions are 5m. So... let x = number of meters past 5.

(5 + x)^2 = 120m^2

3) I honestly am not sure what this one is saying.

1) LET THE WIDTH BE X

LENGTH = 2X - 3

AREA = LENGTH x WIDTH = ( 2X -3) ( X) = 2X^2 - 3X = 54

2X^2 -3X -54 = 0
2X^2 -12X + 9X -54 = 0

2X ( X - 6) + 9( X-6) = 0

( 2X +9) ( X-6) = 0

X = 6

answer WIDTH = 6 CM LENGTH = 9CM

2)