A worker receives an annual wage of $20,000, which he always deposits into a savings account at the end of the year. By the end of the third year (when he

We can solve this :

20000(1 + r)^2   + 20000(1 + r)   + 20000   = 66200     subtract  66200 from each side

20000 (1 + r)^2 + 20000 (1 + r)  - 46200     =  0         let  1 + r  = x

20000x^2 + 20000x  - 46200  = 0       divide through by 20000

x^2 + x - 231/100  =0         add 231/10  to both sides and completet the square on x

x^2 + x + 1/4 =   231/100 + 1/4      factor and simplify

(x + 1/2) ^2   =  256/100      take the positive square root of both sides

x + 1/2 =  sqrt(256/10)

x + 1/2  = 16/10

x + 1/ 2 =    8/5

x = [ 8/5 - 1/2 ]  =     11/10   = 1.1

So.....  1 + r  = 1.10      subtract 1 from both sides

r = .10   =  10%

66200=20000((1 + R)^3 - 1) / R), solve for R
Solve for R:
66200 = (20000 ((R+1)^3-1))/R

66200 = (20000 ((R+1)^3-1))/R is equivalent to (20000 ((R+1)^3-1))/R = 66200:
(20000 ((R+1)^3-1))/R = 66200

Multiply both sides by R:
20000 ((R+1)^3-1) = 66200 R

Expand out terms of the left hand side:
20000 R^3+60000 R^2+60000 R = 66200 R

Subtract 66200 R from both sides:
20000 R^3+60000 R^2-6200 R = 0

The left hand side factors into a product with four terms:
200 R (10 R-1) (10 R+31) = 0

Divide both sides by 200:
R (10 R-1) (10 R+31) = 0

Split into three equations:
R = 0 or 10 R-1 = 0 or 10 R+31 = 0

Add 1 to both sides:
R = 0 or 10 R = 1 or 10 R+31 = 0

Divide both sides by 10:
R = 0 or R = 1/10 or 10 R+31 = 0

Subtract 31 from both sides:
R = 0 or R = 1/10 or 10 R = -31

Divide both sides by 10:
R = 0 or R = 1/10 or R = -31/10

(20000 ((R+1)^3-1))/R => -(20000 ((1-31/10)^3-1))/(31/10)  =  66200:
So this solution is correct

(20000 ((R+1)^3-1))/R => 20000/0 ((1+0)^3-1)  =  (undefined):
So this solution is incorrect

(20000 ((R+1)^3-1))/R => (20000 ((1+1/10)^3-1))/(1/10)  =  66200:
So this solution is correct

The solutions are:
Answer: |  R = 1/10 or R = -31/10                        R=1/10 x 100 =10%