The quadratic 2x^2 - 6x + 8 can be written in the form a(x + b)^2 + c, where a, b, c and are constants. What is a + b + c?
\(2x^2 - 6x + 8 = 2(x^2 - 3x) + 8=2\left(x^2 - 3x + \dfrac94\right) +\dfrac72 = 2\left(x - \dfrac32\right)^2 + \dfrac72\)
Now you should be able to determine the values of a, b, and c.
