How do you solve #0=x^2-3x-6# by completing the square?

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Answer 1 · 2015-05-17T07:33:03

#0 = x^2-3x-6 = (x^2-3x+9/4)-9/4-6#

#=(x-3/2)^2-33/4#

In general:

#ax^2+bx+c = a(x+b/(2a))^2 + (c-b^2/(4a))#

Adding #33/4# to both sides we get

#(x-3/2)^2 = 33/4#

So #x-3/2 = +-sqrt(33/4) = +-sqrt(33)/2#

Adding #3/2# to both sides we get

#x = 3/2+-sqrt(33)/2 = (3+-sqrt(33))/2#