The general vertex form is

#color(white)("XXX")y=color(green)m(x-color(red)a)^2+color(blue)b#

with vertex at #(color(red)a,color(blue)b)#

Given

#color(white)("XXX")y=2x^2+8x+4#

Extract the #color(green)m# factor (in this case #color(green)2# from the first two terms

#color(white)("XXX")y=color(green)2(x^2+4x)+4#

Supposing that the #(x^2+4x)# are part of a squared binomial #(x-color(red)a)^2=(x^2-2color(red)ax+color(red)a^2)#

then #-2color(red)ax# must be equal to #4x#

#rarr color(red)a=4/(-2)=color(red)(-2)#

and in order to "complete the square" an extra #color(red)a^2=(-2)^2=color(magenta)4# will need to be added inside the brackets to the #(x^2+4x)# we already have.

Note that if we insert this #color(magenta)(+4)# because of the factor #color(green)m=color(green)2#

we will really be adding #color(green)2 xx color(magenta)4=color(brown)8# to the expression.

To maintain equality if we add #color(brown)8# we will also need to subtract it:

#color(white)("XXX")y=color(green)2(x^2+4x+color(magenta)4)+4color(brown)(-8)#

Simplifying

#color(white)("XXX")y=color(green)2(x+2)^2-4#

Adjusting to match the sign requirements of the *vertex form*:

#color(white)("XXX")y=color(green)2(x-color(red)(""(-2)))^2+color(blue)(""(-4))#

The graph below of the original equation supports this result.

graph{2x^2+8x+4 [-8.386, 2.71, -5.243, 0.304]}