Write 32c + 1 as (32c) · (3) or as 3·32c
This problem becomes: 3·32c = 28·3c - 9
which is: 3·32c - 28·3c + 9 = 0
I'm going to replace 3c with x and 32c with x²:
---> 3x² - 28x + 9 = 0
---> Factor: (3x - 1)(x - 9) = 0
Which means: 3x - 1 = 0 or x - 9 = 0
---> x = 1/3 or x = 9
But x = 3c ---> 3c = 1/3 or 3c = 9
---> 3c = 3-1 or 3c = 32
So: c = -1 or c = 2
Write 32c + 1 as (32c) · (3) or as 3·32c
This problem becomes: 3·32c = 28·3c - 9
which is: 3·32c - 28·3c + 9 = 0
I'm going to replace 3c with x and 32c with x²:
---> 3x² - 28x + 9 = 0
---> Factor: (3x - 1)(x - 9) = 0
Which means: 3x - 1 = 0 or x - 9 = 0
---> x = 1/3 or x = 9
But x = 3c ---> 3c = 1/3 or 3c = 9
---> 3c = 3-1 or 3c = 32
So: c = -1 or c = 2