How do you solve the quadratic #y^2+2y+1=0# using any method? Precalculus Linear and Quadratic Functions Completing the Square 1 Answer ali ergin Aug 11, 2016 #y=-1# Explanation: #y^2+2y+1=0# #"Solution -1"# #y^2+2y+1=(y+1)^2# #(y+1)^2=0# #y=-1# #"Solution-2" # #Delta=sqrt(4-4*1*1)=0# #y_("1,2")=(-b±Delta)/(2*a)# #"So "Delta =0 " There is one root"# #y=(-2±0)/(2*1)# #y=-1# 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 10528 views around the world You can reuse this answer Creative Commons License