Question 131950
If the width of the sidewalk is x, then the outside dimensions of the area covered by the factory AND the sidewalk must be {{{60 + 2x}}} and {{{80 + 2x}}}.  See figure:


{{{drawing(600,600,0,10,0,12,
rectangle(1,1,9,11),
rectangle(2,2,8,10),
locate(1.5,6,x),
locate(8.5,6,x),
locate(5,1.5,x),
locate(5,10.5,x),
locate(5,2.5,60),
locate(2.5,6,80)
)}}}


The area covered by the factory is {{{60*80=4800}}} square meters


The area covered by the factory AND the sidewalk is {{{(60+2x)(80+2x)=2*4800=9600}}} square meters.


Expand and collect terms:
{{{4800+120x+160x+4x^2=9600}}}
{{{4800+280x+4x^2=9600}}}


Add -9600 to both sides and put in standard form:
{{{4x^2+280x-4800=0}}}


Divide through by 4:
{{{x^2+70x-1200=0}}}


Complete the square:
{{{x^2+70x=1200}}}
{{{x^2+70x+35^2=1200+35^2=1200+1225=2425}}}
{{{(x+35)^2=2425}}}
{{{x=-35+-sqrt(2425)}}}, but {{{2425=25*97}}} so,


{{{x=-35+5sqrt(97)}}} or {{{x=-35-5sqrt(97)}}}.  Second root is negative, but we are trying to determine a length measure, so discard the second root as extraneous.


{{{x=-35+5sqrt(97)}}}, to the nearest tenth, is 14.2 meters.  I'll let you verify the arithmetic.