How would you solve (3a-4) ^2?

According to my teacher the answer is (3 (a-1)) ^2 = 9a^2 - 18 a + 9

(3a - 4) ^2 =

(3a - 4) (3a - 4) =

3a (3a - 4) - 4 (3a - 4) =

9a^2 - 12a - 12a + 16 =

9a^2 - 24a + 16 < = = I am not getting what ur teacher got

is the problem written correctly?

Because (3a - 3) ^2 = 9a^2 - 18a + 9

Well (3a - 4) ^2 is equivalent to (3a - 4) * (3a - 4)

How I would solve that is by firstly multiplying (3a - 4) by 3a (first part of second bracket) then adding that to (3a - 4) * - 4 (second part of second bracket)

That would equal

(9a^2 - 12a) + (-12a + 16)

= 9a^2 - 12a - 12a + 16

= 9a^2 - 24a + 16

Sorry if my answer is rubbish