Question

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

a = 1, a = –1

a = 5, a = –1

a = 5, a = 1

Answer 1

   
$\displaystyle\\ 15(2a - 2) = 5(a^2 -1) \\ \\ 30a - 30 = 5a^2-5\\\\ -5a^2 + 30a -30 + 5 =0\\\\ -5a^2 + 30a -25 =0 ~~~~~\Big| \times (-1) \\\\ 5a^2 - 30a +25 = 0~~~~~\Big| : 5 \\\\ a^2 - 6a +5 = 0 \\ \\ a_{12}= \frac{-b\pm \sqrt{b^2-4ac}}{2a}= \frac{6\pm \sqrt{36-20}}{2}= \frac{6\pm \sqrt{16}}{2}=\frac{6\pm 4}{2}= \boxed{3\pm 2} \\ \\ a_1 = 3+ 2 = \boxed{5}\\\\ a_2 = 3-2 = \boxed{1}$

Answer 2

Answer:

c

Step-by-step explanation:

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