Help with factoring trinomials ax(2)+bx+c word problems

I kinda get the regular problems, but the word problems just completely stump me. If anyone could help, that'd be great :)

1) The difference of the squares of two consecutive odd integers is 24. Find the integers.

2) The length of a garden is 20 yards greater than its width. The area is 300 square yards. What are the dimensions?

1) let x be first odd number
x+2 be the second

so (x+2)^2-x^2 = 24
x^2+4x+4-x^2 = 24
4x +4 = 24
4x = 20
x = 5

first odd number = 5
second = 7

to check: 7^2-5^2 =49-25+24 :))

2) let x be the width
x+20 be the length
area =300

area = length x width
300 = (x+20)(x)
300 = x^2+20x
x^2+20x-300=0 ...use quadratic or factor out like this:
(x +30) (x-10) =0
then x+30 = 0 x - 10=0
x = -30 x =10

and it's obviously both positive cause it's area. so dimensions are 30 and 10 yards

1. let x be the largest odd integer
let x-2 be the smallest odd integer.

the difference of the two odd integer is 24 so
(x)^2 - (x-2)^2=24
x^2-x^2+4x-4 = 24 cancel x^2
4x-4=24 transpose -4 to the right side
4x=24+4
4x=28 divide both sides by 4
x=7

substitute 7 to x-2
it gives 5
the integers are 5 and 7.



2. let L be the length
let W be the width

L = 20+W
W = W

formula for the rectangle is
A = L * W
(20+W)*(W) = 300 sq.yards
20W + W^2 = 300 sq. yards transpose 300 to the left side.

W^2 + 20W -300=0 by quadratic factoring.
it gives
(W+30)(W-10)=0
W = -30 , 10 but negative is not valid the width is 10

the dimension is 30yards X 10 yards

(2n+1)^2 - (2n-1)^2 = 24
8n = 24
n = 3
Numbers 7 and 5
49-25 = 24

x(x+20) = 300
x^2 + 20 x + 100 = 400
(x+10)^2 = 20^2
x+10 = 20 ( only + 10)
x = 10

1) (x+2)^2 -x^2 =24
x^2 +4x +4 -x^2 = 24
4x =20
x= 5
x+2= 7

2)Area = l x w = 300yds^2
l and l-20
l x(1+20)
l^2 +20 l =300
l^2 +20l -300 =0
(l +30)(l-10)
l =30
w= 10