How do you solve #x^2-6x-11=0# by completing the square?

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Answer 1 · 2016-08-13T23:43:09

#x = 3+-2sqrt(5)#

Move the constant term to the RHS. Take half the #x# coefficient. Square it and add to both sides.

#x^2 - 6x + (-3)^2 = 11 + (-3)^2#

Can simplify this to

#(x-3)^2 = 20#

Take square roots of both sides

#x - 3 = +-sqrt(20) = +-2sqrt(5)#

#x = 3+-2sqrt(5)#