The vertex form:
[tex]y=a(x-h)^2+k\\\\(h,\ k)-vertex[/tex]
The axis of symetry is x = h.
[tex]y=ax^2+bx+c\to h=\dfrac{-b}{2a}[/tex]
We have
[tex]y=2x^2+12x+16\to a=2,\ b=12,\ c=16[/tex]
Substitute:
[tex]h=\dfrac{-12}{(2)(2)}=\dfrac{-12}{4}=-3[/tex]
Answer: x = -3
