The number that should be added to both sides is 16 because its half of b, which is then squared. You then get (x+4)^2=20, which you then square root both sides and get x+4=+-20 and then subtract the 4 to get x=-4+-20
for [tex]ax^2+bx+c[/tex]
if a=1, and [tex]c=(\frac{b}{2})^2[/tex], then it comlpetes the square
[tex]x^2+12x[/tex]
a=1, b=12
since a=1, c=[tex](\frac{b}{2})^2=(\frac{12}{2})^2=(6^2)=36[/tex]
add 36 to both sides of the nonexistant equation to copmlete square
