5 cm less than twice the width:
[tex]l = 2w - 5[/tex]
[tex]A = lw[/tex]
[tex]33 = lw [/tex]
[tex]33 = (2w - 5)(w)[/tex]
[tex]33 = 2w^2 - 5w [/tex]
[tex]2w^2 - 5w - 33 = 0
[/tex]
[tex]w = 5.5, -3[/tex]
-3 is an extraneous solution, so the only answer is [tex]w = 5.5[/tex]
[tex] 33 = 5.5l[/tex]
[tex] \frac {33}{5.5} = ( \frac{5.5}{5.5})(l)
[/tex]
[tex] 6 = l [/tex]
The dimensions of the rectangle are a length of 6, and a width of 5.5
