Question 887008
Another one much like this was asked not too long ago (today).  Try this instead.


x and y are rectangle dimensions.  Assume they are unknown but perimeter p and area A are known as given.


Find x and y.
{{{2x+2y=p}}} and {{{xy=A}}}.
Maybe p is an even number, and maybe it is not.


{{{2y=p-2x}}}
{{{y=(p-2x)/2}}}
and substituting into A,
{{{x(p-2x)/2=A}}}
{{{x(p-2x)=2A}}}
{{{px-2x^2=2A}}}
{{{-2x^2+px-2A=0}}}
{{{highlight_green(2x^2-px+2A=0)}}}, and in practice not doing everything purely symbolically, maybe this is factorable
and maybe it is not factorable.  Continuing, using the general solution for a quadratic equation
{{{x=(p+- sqrt(p^2-4*2*2A))/(2*2)}}}
{{{highlight(x=(p+- sqrt(p^2-16A))/4)}}}.
One of these will makes sense and the other will not.  Use the meaningful one.
Now just use the found x value to get the y value which corresponds.