How do I convert $y= 2x^{2} + 16x$ into the vertex form (i.e. $y=a(x-h)^{2}+k$)? I tried looking up the "process" of solving that equation, but I couldn't really find the exact way to solve it.
Isolating the $2$ from $2x^2$ might be one of the way, but I couldn't exactly find out what I would have to do after that.
Thanks for helping me.

While Dr. Sonnhard Graubner's answer is valid, I'd like to present a more intuitive approach.
Recall: the vertex form of a parabola is given by $y = a(x - h)^2 + k$, for vertex $(h,k)$. For the sake of argument, we can expand that form by foiling the squared term:
$$y = ax^2 - 2hax + ah^2 + k$$
We seek to write $y = 2x^2 + 16x$ in this form. Notice, however, that to generate the same parabola, we will need constants $a,h,k$ such that the two equations are equal. That means we set them equal to each other:
$$2x^2 + 16x = ax^2 - 2hax + ah^2 + k$$
In the interest of clarifying my next step, I will add some extra terms and parentheses:
$$(2)x^2 + (16)x + (0) = (a)x^2 + (-2ha)x + (ah^2 + k)$$
What would it mean for these two polynomials to be equal? Well, the constant terms would equal, the coefficients of the linear term $x$ would be equal, and the coefficients of the quadratic term $x^2$ would be equal. That is to say, we would have three equations:
$$\begin{align}
2 &= a \\
16 &= -2ha \\
0 &= ah^2 + k \\
\end{align}$$
The first equation outright gives us $a = 2$. 
Plug that into the second equation and thus $16 = -4h$. Solve for $h$ and you get $h = -4$.
Plug both into the third equation and you get $0 = 32 + k$. Thus, $k = -32$.
Now we just substitute the $a,h,k$ we found into the vertex form:
$$y = a(x - h)^2 + k = 2(x - (-4)^2 + (-32) = 2(x+4)^2 - 32$$
