It's a method called "completing the square". We add/subtract an appropriate constant term that allows us to condense three terms into a squared binomial, and leaves a constant remainder outside the squared term.
In this case, we first factor out -16 from the first two terms:
[tex]-16t^2+22t+4=-16\left(t^2-\dfrac{11}8t\right)+4[/tex]
When we square a binomial like [tex](x-y)^2[/tex], we end up with the expansion [tex]x^2-2xy+y^2[/tex], so that in the quadratic above we need [tex]-2xy=-\dfrac{11}8t[/tex] with [tex]x=t[/tex] and [tex]y=-\dfrac{11}{16}[/tex]. Then the constant term we have to add/subtract will be [tex]y^2=\left(\dfrac{11}{16}\right)^2[/tex]:
[tex]-16t^2+22t+4=-16\left(t^2-\dfrac{11}8t+\left(\dfrac{11}{16}\right)^2-\left(\dfrac{11}{16}\right)^2\right)+4[/tex]
Now the first three terms within the parentheses can be condensed to get
[tex]-16t^2+22t+4=-16\left(\left(t-\dfrac{11}{16}\right)^2-\left(\dfrac{11}{16}\right)^2\right)+4[/tex]
and all that's left to do is distribute a factor of -16 and combine the remaining constant term:
[tex]-16t^2+22t+4=-16\left(t-\dfrac{11}{16}\right)^2+\dfrac{11^2}{16}+4[/tex]
[tex]-16t^2+22t+4=-16\left(t-\dfrac{11}{16}\right)^2+\dfrac{185}{16}[/tex]
Then the vertex of the parabola defined by this expression occurs at [tex]\left(\dfrac{11}{16},\dfrac{185}{16}\right)[/tex].
