Spectral Line Calculation

[Nova Akhavan 2F]

For the problem: "A violet line is observed at 434.0 nm in the spectrum of atomic hydrogen. Determine the values of n for the beginning and ending energy levels of the electron during the emission of energy that leads to this spectral line."

One of the steps in finding the solution for n2 is:

1) 6.908×10^14 Hz=(3.29×10^15 Hz)(1/2^2−1/n2^2)

2) 6.908×10^14=8.23×10^14−(3.29×1015)/n2^2

What is the process that takes step 1 to step 2?

[Jennifer Huynh 3I]

Given the information you provided, I think step 2 is a simplification of step 1. Step 2 distributes 3.29×10^15 Hz into each term in (1/2^2−1/n2^2).
(3.29×10^15 Hz)(1/2^2−1/n2^2)
(3.29×10^15 Hz)*(1/2^2) - (3.29×10^15 Hz)(1/n2^2)
8.23×10^14−(3.29×1015)/n2^2

And to clarify, we know that n1 = 2 because 434.0 nm corresponds to a wavelength in the Balmer series where an electron transitions from n>= 3 to n = 2 (final state).

From here, we can solve for n2.