Solve b^2 + 2b + 1 = 324.
The number 121 in, say base 10, takes the form
1*10^2 + 2*10^1 + 1* 10^0 = 1*100 + 2*10 + 1 = 121
In a base "b" the number 121 represents:
1*b^2 + 2*b^1 + 1*b^0 = b^2 + 2b + 1
(So 121 in base 3 represents the decimal number 3^2 + 2*3 + 1 = 9 + 6 + 1 = 16.)
For this number to be equal to the decimal number 324 we must have
b^2 + 2b + 1 = 324
-Dan