The length of a rectangle is 59 inches greater than twice the width. if the diagonal is 2 inches more than the length, find the dimensions of the rectangle.

L = 59+2w
diagonal = 2+l

diagonal^2 = l^2 +w^2

(2+l)^2 = l^2 +w^2

4 +4l +l^2 = l^2 +w^2

4+4l =w^2

l = 59+2w

4+4(59+2w)=w^2

4+236+8w=w^2

w^2 -8w -240 =0

D = 64 +960 = 1024

w_1,2 = (8+/-sqrt1024)/2 = -12 and 20

l=59+2*20=59+40=99

l=99
w=20

hope helped 

It should be a or d i’m not sure