#(x+2)^2 = x^2+4x+4#

#5(x+2)^2 = 5x^2+20x+20#

So

#f(x) = 5x^2+20x+4#

#= 5x^2+20x+20-16#

#= 5(x+2)^2-16#

The minimum value of #f(x)# will occur when #x=-2#

#f(-2) = 0-16 = -16#

Hence the range of #f(x)# is #[-16, oo)#

More explicitly, let #y = f(x)#, then:

#y = 5(x+2)^2 - 16#

Add #16# to both sides to get:

#y + 16 = 5(x+2)^2#

Divide both sides by #5# to get:

#(x+2)^2 = (y+16)/5#

Then

#x+2 = +-sqrt((y+16)/5)#

Subtract #2# from both sides to get:

#x = -2+-sqrt((y+16)/5)#

The square root will only be defined when #y >= -16#, but for any value of #y in [-16, oo)#, this formula gives us one or two values of #x# such that #f(x) = y#.