條數唔識計！　求R？ 請詳解！Please！?

real solutions
r = -2.2935
r = -2.25849
r = -2.19123
r = -2.16701
r = -1.95285

complex solutions
r = -1.24479 + 1.09986i
r = -1.24214 - 1.10414i
r = -1.06522 + 1.12115i
r = -1.06518 - 1.12102i
r = -0.886753 - 1.11496i

Sum n=0 to 39 of 900(1+r)^n
=900+900(1+r)+900(1+r)^2+...+900(1+r)^39
=900[(1+r)^40-1]/(1+r-1) using sum of geometric progression with common ratio r
=900[(1+r)^40-1]/(1+r-1)
=900(x^40-1)/(x-1) ............ setting x=1+r

Therefore the given equation can be written
f(x)=60662.30(x-1)-900x^40
is equivalent to solving
f(x)=60662.30x-900x^40-59762.3

Using newton's method, where next approximation x1 from x0 given by
x1=x0-f(x0)/f'(0)
and f'(x)=60662.30-36000x^40
We will attempt an approximation x=1.02
x0=1.02
x1=1.02-f(1.02)/f'(1.02)=1.027297
x2=1.027297-f(1.027297)/f'(1.027297)=1.02523722
x3=1.02523722-f(1.02523722)/f'(1.02523722)=1.02500294512
x4=1.02500294512-f(1.02500294512)/f'(1.02500294512)=1.025000001812965
x5=1.025000001812965-f(1.025000001812965)/f'(1.025000001812965)=1.025000001812965
So we get x=1.0250000
R=x=1+r=1.0250000