Find the length of sides of rectangle

Hi i'm kinda confused in this question.
A rectangle has perimeter 23 cm. Its area is 33cm^2. Determine the dimensions of the rectangle?
Any help is appreciated
Thank You

Perimeter eqn: 2b + 2h = 23
Area eqn: bh = 33

Solve for b in the perimeter eqn. b = 23/2 - h
Plug into area eqn

(23/2 - h)h = 33
(23/2)h - h^2 = 33
h^2 - (23/2)h + 33 = 0
Solving the quadratic equation, h = 6.

So b(6) = 33, b = 5.5

Answer: The rectangle has a base of 5.5 cm and a height of 6 cm.

Length = L
Width = W
Perimeter, P = 23 cm.
Area, A = 33 sq. cm.

2L + 2W = P
2L + 2W = 23
L + W = 23 / 2
L + W = 11.5
W = 11.5 - L

LW = A
LW = 33
L(11.5 - L) = 33
11.5L - L² = 33
L² - 11.5L = - 33
L² - 11.5L + 33.0625 = - 33 + 33.0625
(L - 5.75)² = 0.0625
L - 5.75 = √0.0625
L - 5.75 = ± 0.25
L = 5.75 ± 0.25

If L = 5.75 + 0.25,
L = 6

and

W = 11.5 - 6
W = 5.5


If L = 5.75 - 0.25,
L = 5.5

and

W = 11.5 - 5.5
W = 6

The dimensions are 5.5 cm. x 6 cm.
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w^2 - 11.5w + 33 = 0
( - 5.75 + k ) ( - 5.75 - k ) = 33
33.0625 - k^2 = 33
0.0625 = k^2
.25 = k

- 5.5 and - 6 are the numbers

( w - 5.5 ) ( w - 6 ) = 0
w = 6 or w = 5.5

Therefore, the length is 6 cm and the width is 5.5 cm.