Can you find the dimensions

A rectangle has a perimeter of 23 cm and an area of 33 cm^2. What's the dimensions?
You cannot guess and check. We are on quadratics so I believe I need an equation.

WL = 33 => W = 33/L
2L + 2W = 23
2L + 66/L = 23
2L^2 - 23L + 66 = 0
(L - 6)(2L - 11) = 0
L = 6 , 11/2 => reject 11/2
L = 6 cm
W = 33/6 = 11/2 cm

Length = x.
Width = y.
Perimeter = 2(x + y) = 23 cm ---> x + y = 23/2 cm.
Area = x*y = 33 cm^2.
You've got the sum (S) and the product (P), so:
t^2 - St + P = 0
t^2 - (23/2)t + 33 = 0 (Multiply the entire equation by 2)
2t^2 - 23t + 66 = 0
(2t - 11)(t - 6) = 0
Then:
2t - 11 = 0 ---> 2t = 11 ---> t1 = 11/2 = 5.5 cm.
t - 6 = 0 ---> t2 = 6 cm.
The width s less than the length, so t1 s the width and t2 s the length.
Then, the width is 5.5 cm and the length is 6 cm.

GOD bless you...