Question 914378
("shorter sides", length to find, so these would not be along the wall).


Dimensions x and y.
Fencing quantity is 60 meters, and area to enclose is 448 sq. m's.


Assuming y is along the wall, and another length y is opposite the wall,
x would be one of the shorter sides.


2x+y=60 to account for the length of fencing
and
xy=448 for the area.


{{{highlight_green(system(2x+y=60,xy=448))}}}


Substitute for y in the area equation:
{{{x(60-2x)=448}}}
simplify and solve first for x.
{{{x(30-x)=224}}}
{{{-x^2+30x-224=0}}}
{{{highlight_green(x^2-30x+224=0)}}}


Trying to factor that may be a chore, but the discriminant is {{{30^2-4*224=4}}}, a perfect square.
{{{x=(30+-sqrt(4))/2}}}
{{{x=(32/2)}}}  or {{{x=28/2}}}
{{{x=16}}} or {{{x=14}}}


Earlier found y=60-2x.
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If x=16 then y=28.
If x=14 then y=32.
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