Alright i came upon this problem while tutoring today and it just messed me up.
$$a^2 - 8a + 15 = 0$$
we start by adding 1 to both sides and get
$$a^2 - 8a + 16 = 1$$
then change the left side to
$$(a-4)^2 = 1$$
then take the square root of both sides to get
$$\pm (a - 4) = \pm 1$$
then solve for $a$ in $a - 4 = 1$ to get $$a = 5$$ and $-(a - 4) = -1$ to
get $$a = 5$$
making the answer only one root which is impossible here.
The right answer should be $(5,3)$ but I can't seem to understand how to get $3$. What am I forgetting? Thanks for the help