How do you solve #x^2+1.4x=1.2# by completing the square? Precalculus Linear and Quadratic Functions Completing the Square 1 Answer Barney V. Dec 27, 2016 #x=-0.6 or x=-2# Explanation: #x^2+1.4x=1.2# #(x^2+1.4x-1.2=0# #(x-0.6)(x+2)# #x-0.6=0# #x=0.6# #x+2=0# #x=-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 1377 views around the world You can reuse this answer Creative Commons License