Question 397939
One side is {{{x}}} ft
The other side is {{{x+3}}} ft
{{{A = x*(x+3)}}}
given:
{{{A = 40}}} ft2
------------
{{{40 = x*(x+3)}}}
{{{40 = x^2 + 3x}}}
{{{x^2 + 3x - 40 = 0}}}
Now you can use the quadratic formula:
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} 
{{{a  =1}}}
{{{b = 3}}}
{{{c = -40}}}
{{{x = (-3 +- sqrt( 3^2-4*1*(-40) ))/(2*1) }}}
{{{x = (-3 +- sqrt( 9 + 160 ))/2 }}} 
{{{x = (-3 +- sqrt( 169 ))/2 }}} 
{{{x = (-3 + 13)/2}}}
{{{x = 5}}} ft answer
and
{{{x = (-3 - 13)/2}}}
{{{x = -8}}} (can't use the negative root)
also,
{{{x + 3 = 8}}}
{{{A = 5*8}}}
{{{A = 40}}} ft2