The length of a soccer pitch is 20 m less than twice its width. the area of the pitch is 6000m^2. find its dimensions.

X=width of a coccer pitch
2x-20=length of a soccer pitch
Area (rectangule)=length x width
We suggest this equation:
x(2x-20)=6000
2x²-20x=6000
2x²-20x-6000=0
x²-10x-3000=0
We solve this quadratic equation:

x=[10⁺₋√(100-4*1*-3000)]/2=[10⁺₋√(100+12000)]/2=
=(10⁺₋110)/2
we have two solutions:
x₁=(10-110)/2=-50, invalid solution.
x₂=(10+110)/2=60

x=60
2x-20=2(60)-20=120-20=100

Solution: the length is 100 m, and the width is 60 m.

To check:
Area=100 m*60 m=6000 m²
The twice of width is =2(60 m)=120 m,
20 m less than twice its width is: 120 m-20 m=100 m=the length.

D. i think it is the savings