Question 883362
x and y dimensions.
Perimeter,2x+2y=40 simplifiable to x+y=20.
Area, xy=84.


Use substitution.
{{{y=20-x}}}.
{{{x(20-x)=84}}}.
{{{-x^2+20x-84=0}}}
{{{highlight_green(x^2-20x+84=0)}}}.


Discriminant:  {{{(-20)^2-4*84=64=8^2}}}
Continue in the general solution for a quadratic equation:
{{{x=(20+8)/2=14}}}, guessing that the PLUS form should work and not the MINUS form.


{{{y=20-14=6}}}; but actually, either PLUS or MINUS form would work.  One way would give 14 as found, and the other would give 6.

<b>The dimensions are 14 and 6</b>