How can I solve, #2x^2+6x+2=0# by completing the square? Precalculus Linear and Quadratic Functions Completing the Square 1 Answer Martin C. Jul 9, 2018 #x_1=(-3-sqrt(5))/2 or x_2=(-3+sqrt(5))/2# Explanation: #2x^2+6x+2=0|:2# #x^2+3x+1=0# #(x+3/2)^2-9/4+1=0|+5/4# #x+3/2=+-sqrt(5)/2|-3/2# #x=(-3+-sqrt(5))/2# #x_1=(-3-sqrt(5))/2 or x_2=(-3+sqrt(5))/2# 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 3104 views around the world You can reuse this answer Creative Commons License