Given #r^2-9r-38=-9#
First add 9 to both sides to get:
#r^2-9r-29=0#
Then
#0 = r^2-9r-29#
#= r^2-9r+81/4-81/4-29#
#= (r-9/2)^2-(81/4+29)#
#= (r-9/2)^2-(81/4+116/4)#
#= (r-9/2)^2-197/4#
Adding #197/4# to both ends of this equation, we find:
#(r-9/2)^2 = 197/4#
So:
#r-9/2 = +-sqrt(197/4) = +-sqrt(197)/sqrt(4) = +-sqrt(197)/2#
Add #9/2# to both sides to get:
#r = 9/2+-sqrt(197)/2 = (9+-sqrt(197))/2#
In general:
#ar^2+br+c = a(r+b/(2a))^2 + (c - b^2/(4a))#
which is zero when
#a(r+b/(2a))^2 = (b^2/(4a) - c) = (b^2 - 4ac)/(4a)#
and hence
#(r+b/(2a))^2 = (b^2 - 4ac)/(4a^2)#
so
#r+b/(2a) = +-sqrt((b^2 - 4ac)/(4a^2)) = +-sqrt(b^2 - 4ac)/sqrt(4a^2)#
#= +-sqrt(b^2-4ac)/(2a)#
Subtracting #b/(2a)# from both sides we get:
#r = -b/(2a)+-sqrt(b^2-4ac)/(2a) = (-b +- sqrt(b^2-4ac))/(2a)#
Does that look familiar?
