f(x) = (x - 3)^2 + 5
f(x) = x^2 - 6x + 9 + 5
f(x) = x^2 - 6x + 14
f'(x) = 2x - 6
f'(2) = 2*2 - 6
f'(2) = 4 - 6
f'(2) = -2
So the slope of the tangent line is -2, so the tangent line is y = -2x + b, which passes through (2,6), therefore:
6 = -2(2) + b
6 = -4 + b
b = 6 + 4
b = 10
So the tangent line is y = -2x + 10