The problem is a squared-2a-8=0 I do not know how to do this because I was absent. My teacher told me the answer, which is (4,-2), but I do not get how to solve it! Please help

In any quadratic, before you complete the square, you first divide through by the leading coefficient to make it 1. In your case of a^{2} - 2a - 8 = 0, the leading coefficient is already 1.

Now you can complete the square. If the linear coefficient is 2r then you want to make the constant coefficient r^{2}. Why? This is because of the fact that a^{2} + 2ra + r^{2} = (a + r)^{2}

In your case, 2r = -2 so r = -2/2 = -1. Therefore we want the constant coefficient to be r^{2} = (-1)^{2} = 1.

a^{2} - 2a - 8 = 0

a^{2} - 2a + 1 = 9

(a - 1)^{2} = 3^{2}

a - 1 = ±3

a = 1 ± 3

a = 4 or a = -2