Question

a parking lot has a rectangular area of 40,000yd^2. The length is 200 yd more than twice the width. What are the dimensions of the lot? So far I have width=x length=2x+200 so A=lw so 40,000=(2x+200)(x)

Answer 1

then i took 2x^2 +200x=40,000 everything after that is a mess

Answer 2

answer should be 100x400 how do i get there?

Answer 3

Set it equal to 0:
\[2x^{2} + 200x - 40000 = 0\]
Then factor:
\[(2x - 200)(x+200)\]

Answer 4

\[x=100\]\[(x+200)(x-100)=0\]\[x^2+100x-20000=0\]