Since the coefficient of x^2 is positive, this quadratic is a parabola in the shape of a U, hence has a minimum.
We want to end up with the form (x-h)^2 + c. Since (x-h)^2>=0, this form shows that the minimum is achieved when x=h.
Completing the square will put the quadratic in the desired form. Note that:
Comparing this with the given form, we must have -8=-2h, or h=4. But we are missing h^2=4^2=16. We can add the missing 16 and subtract it elsewhere without changing the quadratic.
x^2-8x+16 + (16-4) = (x-4)^2 + 12
Now we know that at x=4 the quadratic has a minimum and that the minimum is 12.