The solutions to 15(2a – 2) = 5(a2 – 1) are:

Treat a2 as a². Expanding gives 30a − 30 = 5a² − 5, which rearranges to 5a² − 30a + 25 = 0. Divide by 5 and factor: (a − 1)(a − 5) = 0, so a = 1 or a = 5. This matches choice C, a = 5 and a = 1.

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