Verified answer

Given 1080.42 = 30 [1 - ( 1 /(1+r)^10 )/r ] + 1000 / (1 + r)^10

Lets just clean up that right hand side first--sheesh

30 [1 - 1 /(r (1+r)^10) ] + 1000 / (1 + r)^10

30 - 30/(r (1 + r)^10) + 1000/(1 + r)^10

30 - 30/(r (1 + r)^10) + (1000r)/( r + (1 + r)^10)

30 + (1000r - 30)/(r ( 1 + r )^10)

Ok lets put the equation back together now

1080.42 = 30 + (1000r - 30)/(r ( 1 + r )^10)

1050.42 = (1000r - 30)/( r ( 1 + r )^10)

1050.42 r ( r + 1) ^10 = 1000r - 30

1050.42 r ( r + 1) ^ 10 - 1000r + 30 = 0

Ok I am NOT expanding that out, especially when there are theoretically 11 roots. I don't know what class this is for but you can do some Newton's method here.

Let f(x) = 1050.42 r ( r + 1) ^10 - 1000r + 30

Either way there is only one real root at r = -1.99658.

Apparently with some patience you can achieve this:

1050.42 (1.99658 + r) (0.00284936 + (-0.00811115 + r) r) (0.386017 +

r (0.392602 + r)) (1.38113 + r (1.38791 + r)) (2.61114 +

r (2.61796 + r)) (3.60623 + r (3.61306 + r)) = 0, which does indeed deliver the one real root. But I got this using a CAS.

r^19=51.42