A rectangular field that is 119 yards in area is to be covered with sod. It is kow that the field is 3 yards longer than twice the width Find the dimensions of the field. I got to the equation 2w2=3w-119=0 this was solved by factoring and I cannot figure ou then next step which was 2w2-14w-17w-199=0 And help in explaning this would be appreciated

Question

A rectangular field that is 119 yards in area is to be covered with sod.  It is kow that the field is 3 yards longer than twice the width  Find the dimensions of the field.  I got to the equation 2w2=3w-119=0  this was solved by factoring and I cannot figure ou then next step which was 2w2-14w-17w-199=0  And help in explaning this would be appreciated

phi · Answer

the length L is 3 longer than 2w,  or L= 3+2w

area is L*w

A= (2w+3)*w 
or
(2w+3)*w = 119

phi · Answer

now put this into standard form
(2w+3)w = 119
2w^2 +3w - 119=0

do you know the quadratic formula to solve for w ?

anonymous · Answer

Let width =w yards
Length=2w+3
w(2w+3)=119
2w^2+3w-119=0
2*-119=-238
17-14=3
17*-14=-238
2w^2+17w-14w-119=0
w(2w+17)-7(2w+17)=0
(2w+17)(w-7)=0
now you can find the values.

anonymous · Answer

Example 1: Rectangle has the length 13 cm and width 8 cm. solve for perimeter of rectangle.
Solution: 
Given that:
Length (l) = 13 cm
Width (w) = 8 cm

Perimeter of the rectangle = 2(l + w) units 
P = 2(13 + 8)
P = 2 (21)
P = 42