How do you solve #y^2-6y-4=0# by completing the square?

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Answer 1 · 2015-04-29T12:15:29

#y^2-6y-4 = 0#

Move the constant term to the right side:
#y^2-6y = 4#

Add the value of #a^2# to both sides where
#2a = -6#
since we want #(y+a)^2#
# = y^2 +2ax +a^2#
# = y^2-6y+a^2#

#a =-3#
#a^2 = 9#

#y^2-6y+9= 4+9#

#(y-3)^2 = 13#

#y-3 = +-sqrt(13)#

#y = 3 +-sqrt(13)#