How do you complete the square #3x^2-6x-1=0#? Precalculus Linear and Quadratic Functions Completing the Square 1 Answer Massimiliano Feb 4, 2015 The answer is: #3(x^2-2x)-1=0rArr3(x^2-2x+1-1)-1=0rArr# #3(x^2-2x+1)-3-1=0rArr3(x-1)^2-4=0rArr# #(x-1)^2=4/3rArrx-1=+-sqrt(4/3)=0rArr# #x=1+-2/sqrt3rArrx=1+-(2sqrt3)/3rArr# #x=(3+-2sqrt3)/3#. Answer link Related questions What does completing the square mean? How do I complete the square? Does completing the square always work? Is completing the square always the best method? Do I need to complete the square if #f(x) = x^2 - 6x + 9#? How do I complete the square if #f(x) = x^2 + 4x - 9#? How do I complete the square if the coefficient of #x^2# is not 1? How do I complete the square if #f(x) = 3x^2 + 12x - 9#? If I know the quadratic formula, why must I also know how to complete the square? How do I use completing the square to describe the graph of #f(x)=30-12x-x^2#? See all questions in Completing the Square Impact of this question 2709 views around the world You can reuse this answer Creative Commons License