Verified answer

first ignore the square root and treat it normally:

f(x)=(2/5)x^2+10x+30

f(x)=(2/5)(x^2+10x+_)+30-(2/5)_

to complete the square, take the middle term, divide it by 2, and square it:

(10/2)^2=5^2=25

so

f(x)=(2/5)(x^2+10x+25)+20

f(x)=(2/5)(x+5)^2+20

now put the square root back

f(x)=((2/5)(x+5)^2+20)^(1/2)

f(x)=√[(2/5)x²+10x+30]

=√{2/5[x²+25x+(12.5)²]+(30-[12.5²][2/5]}

=√[2/5[x+12.5)²-32.5]