P = 2L +2w = 58 ----> L+w = 29 ----> L = 29-w
A = L*w = 204
Now substitute "29-w" for L in Area equation:
(29-w)w = 204
This gives a quadratic
w^2 - 29w + 204 = 0
Use quadratic formula:
[tex]w = \frac{29 \pm \sqrt{29^2 -4(1)(204)}}{2} = \frac{29 \pm 5}{2}[/tex]
w = 12, w = 17
Either value will work since L=29-w, the other number becomes length.
Therefore the final dimensions of the rectangle are 12 by 17.
