Question 861223
Let s = side, {{{s^2}}} = area, {{{4s}}} = perimeter
the area of a square is 77 more than the perimeter
s^2 is 77 more than 4s
{{{s^2=4s+77}}}
Subtract 4s+77 from both sides
{{{s^2-4s-77=0}}}
Use the quadratic formula
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
{{{x = (4 +- sqrt( 4^2-4*1*-77 ))/(2*1) }}}
{{{x = (4 +- sqrt( 16+308 ))/(2) }}}
{{{x = (4 +- sqrt( 324 ))/(2) }}}
{{{x=(4+18)/2}}} OR {{{x=(4-18)/2}}}
{{{x=11}}} OR {{{x=-7}}}
Because the length can't be negative, it's 11.