Find the dimensions of a rectangle with perimeter 100m area of 336

Start by writing out the formulas for perimeter and area:
P=2L+2W                                                                A=L*W

We know that A = L*W = 336 sq meters and that P=2L + 2W = 100m.

Let's solve the first equation for L:  L = 336/W

Subst. this into the formula for perimeter:  2[336/W] + 2W = 100 m (given)

Now all you have to do is to solve this for W:

336/W + W = 50            becomes 336 + W^2 = 50W.

Then W^2 - 50W + 336 = 0.  Using the quadratic formula, I found that 
          50 plus or minus sqrt (1156)         50 plus or minus 34
W =  =  
                                2                                              2

W has to be positive.  Thus, choose W = (50+34) / 2 = 42

Then L = 336 / W, or 336 / 42, or 8.

The dimensions of this rectangle are 8 by 42.

You should check these results.  Does P = 2(8) + 2(42) = 100 ?