Since the coefficient of #x^2# is 1, we can take the coefficient of #x#, the 20, divide it by 2 to get 10, and then square that number to get 100. Now add zero by adding and subtracting 100 to get:

#x^2+20x=(x^2+20x+100)-100#

But the expression #x^2+20x+100# is a perfect square:

#x^2+20x+100=(x+10)^2#.

Hence,

#x^2+20x=(x^2+20x+100)-100=(x+10)^2-100#

One purpose of this procedure is to help you graph the function #f(x)=x^2+20x#. Since #f(x)=(x+10)^2-100# as well, you can see that the minimum value of #f(x)# is #-100# when #x=-10# (note that the coefficient of #(x+10)^2# is 1, which is positive, so the vertex is a minimum rather than a maximum). The vertex of the parabola that is the graph of #f# is at #(x,y)=(-10,-100)#. The graph opens upward, and you can find other things to help you graph it, such as #x# and #y#-intercepts (the #y#-intercept is #f(0)=0# and the #x#-intercepts are #x=0# and #x=-20# since #f(x)=x^2+20x=x(x+20)#.

Can you use the method above to complete the square in the following general case? #f(x)=x^2+bx+c#? Give it a try and see if you can figure out where the vertex of the graph of #f# is after doing so (its coordinates depend on #b# and #c#).