SOLUTION: the length of the rectangle is 5yd more then its width, the area of the rectangle is 88yd^2. find the dimensions of the rectangle

Algebra ->  Rectangles -> SOLUTION: the length of the rectangle is 5yd more then its width, the area of the rectangle is 88yd^2. find the dimensions of the rectangle      Log On


   

Question 358691: the length of the rectangle is 5yd more then its width, the area of the rectangle is 88yd^2. find the dimensions of the rectangle
Answer by checkley77(12844) About Me  (Show Source):
You can put this solution on YOUR website!
L=W+5
LW=88
(W+5)W=88
W^2+5W=88
W^2+5W-88=0
w+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
w=(-5+-sqrt[5^2-4*1*-88])/2*1
w=(-5+-sqrt[25+352])/2
w=(-5+-sqrt377)/2
w=(-5+-19.4165)/2
w=(-5+19.4165)/2
w=14.4165/2
w=7.2 ANS.
w=(-5-19.4165)/2
w=(-24.4165/2
w=-12.2 ans.