Question 122670
First let's draw a picture. We can see that the length of the inner rectangle is  {{{300-2x}}} and the width of the inner rectangle is {{{200-2x}}} since we are taking away {{{2x}}} from both the length and the width.


note: the inner rectangle represents the remaining area


{{{drawing(500,500,-10,7,-7,7,
blue(line(-5,3,5,3)),
blue(line(5,3,5,-3)),
blue(line(5,-3,-5,-3)),
blue(line(-5,-3,-5,3)),
red(line(-6,4,6,4)),
red(line(6,4,6,-4)),
red(line(6,-4,-6,-4)),
red(line(-6,-4,-6,4)),
green(line(-5,3,-5,4)),
green(line(-5,3,-6,3)),
green(line(5,3,5,4)),
green(line(5,3,6,3)),
green(line(-5,-3,-5,-4)),
green(line(-5,-3,-6,-3)),
green(line(5,-3,6,-3)),
green(line(5,-3,5,-4)),
red(locate(-0.5,4.5,Length:300)),
blue(locate(-0.5,3.5,Length:300-2x)),
red(locate(-8,0,Width:200)),
blue(locate(-7.95,1.2,Width:200-2x)),
green(locate(-4.8,3.8,x)),
green(locate(-5.6,3,x))
)}}}



From the drawing, the area of remaining area is (ie the blue rectangle):


{{{A=(300-2x)(200-2x)}}} 




{{{A=(300-2x)(200-2x)}}} Start with the given equation




{{{50000=(300-2x)(200-2x)}}} Plug in A=50,000




{{{50000=4x^2-1000x+60000}}} Foil




{{{0=4x^2-1000x+60000-50000}}} Subtract 50,000 from both sides




{{{0=4x^2-1000x+10000}}} Combine like terms



Let's use the quadratic formula to solve for x:



Starting with the general quadratic


{{{ax^2+bx+c=0}}}


the general solution using the quadratic equation is:


{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a)}}}




So lets solve {{{4*x^2-1000*x+10000=0}}} ( notice {{{a=4}}}, {{{b=-1000}}}, and {{{c=10000}}})





{{{x = (--1000 +- sqrt( (-1000)^2-4*4*10000 ))/(2*4)}}} Plug in a=4, b=-1000, and c=10000




{{{x = (1000 +- sqrt( (-1000)^2-4*4*10000 ))/(2*4)}}} Negate -1000 to get 1000




{{{x = (1000 +- sqrt( 1000000-4*4*10000 ))/(2*4)}}} Square -1000 to get 1000000  




{{{x = (1000 +- sqrt( 1000000+-160000 ))/(2*4)}}} Multiply {{{-4*10000*4}}} to get {{{-160000}}}




{{{x = (1000 +- sqrt( 840000 ))/(2*4)}}} Combine like terms in the radicand (everything under the square root)




{{{x = (1000 +- 200*sqrt(21))/(2*4)}}} Simplify the square root (note: If you need help with simplifying the square root, check out this <a href=http://www.algebra.com/algebra/homework/Radicals/simplifying-square-roots.solver> solver</a>)




{{{x = (1000 +- 200*sqrt(21))/8}}} Multiply 2 and 4 to get 8


So now the expression breaks down into two parts


{{{x = (1000 + 200*sqrt(21))/8}}} or {{{x = (1000 - 200*sqrt(21))/8}}}



Now break up the fraction



{{{x=+1000/8+200*sqrt(21)/8}}} or {{{x=+1000/8-200*sqrt(21)/8}}}



Simplify



{{{x=125+25*sqrt(21)}}} or {{{x=125-25*sqrt(21)}}}



So these expressions approximate to


{{{x=239.564392373896}}} or {{{x=10.435607626104}}}



So our possible solutions are:


{{{x=239.564392373896}}} or {{{x=10.435607626104}}}



However, if you plug in {{{x=239.564392373896}}} into {{{A=(300-2x)(200-2x)}}} , it will give you a negative area, so our only solution is 



{{{x=10.435607626104}}}



So the roadway can be about 10.44 m wide which will leave 50,000 m^2 of field