Question 59822
Let the width of the pool = W and the length of the pool = L. We are told that the length is greater than W so {{{L > W}}}.

Assume the three sides that are fenced in are the two widths and one length, and that's with 60 yards of fencing. Thus, 

(1) {{{2W+L = 60}}}

We also know that the area of the pool = 352. So, 

(2) {{{LW = 352}}}

Subtract {{{2W}}} from both sides of (1) to get {{{L = 60-2W}}}

Replace L with {{{60-2W}}} in (2) to get 

(3) {{{(60-2W)W = 352}}}

Now, multiply {{{60-2W}}} by W and we get {{{60W-2W^2}}}. 
So, (3) becomes {{{60W-2W^2 = 352}}} or 

(4) {{{60W-2W^2-352 = 0}}}

This is a quadratic equation which can also be written as {{{-2W^2+60W-352 = 0}}}. 

We divide both sides by 2 and get {{{-W^2+30W-176 = 0}}}.

The coefficients of this quadratic equation are -1, 30, and -176.

Remember the quadratic equation, solving for w: {{{W = (-b +- sqrt(b^2-4*a*c ))/(2*a) }}}

Replace a, b, and c with -1, 30, and -176 respectively and you get

{{{W = (-(30) +- sqrt(30^2 - 4*(-1)*(-176))) / (2*(-1))}}}

Multiply things out and you get

{{{W = (-30 +- sqrt(900 - 704)) / -2}}} or

{{{W = (-30 +- sqrt(196)) / -2}}}

{{{sqrt(196) = 14}}} so {{{W = (-30 +- 14) / -2}}}

So, W can equal {{{(-30 + 14) / -2}}} which = 8
or W can equal {{{(-30 - 14 ) / -2}}} which = 22

From (1) we know that {{{2W+L = 60}}}. If W = 8 then {{{2*W+L = 60}}} or {{{16+L = 60}}} so {{{L = 44}}}.

If W = 22 then {{{2*W+L = 60}}} or {{{2*22 + L = 60}}} or {{{44+L = 60}}} so {{{L=16}}}. But, L must be > W and 16 is not > 22 so W = 22 is not a solution.

So, L = 44. and from (1) we know that {{{2W+L = 60}}} so {{{2W+44 = 60}}} so {{{2W = 16}}} and {{{W = 8}}}.

To verify our answer, that {{{L = 44}}} and {{{W = 8}}} we plug these values into (1) and (2):

{{{2W+L = 60}}} so {{{2*8 + 44}}} should = 60 and it does.
{{{LW = 352}}} so {{{44*8}}} should = 352 and it does.