How do you solve #y^2 + 12y + 11 = 0#? Algebra Quadratic Equations and Functions Comparing Methods for Solving Quadratics 1 Answer Gerardina C. Jun 5, 2016 #y=-1 or y=-11# Explanation: #y=-6+-sqrt(6^2-11)# #y=-6+-sqrt(25)# #y=-6+-5# #y=-1 or y=-11# Answer link Related questions What are the different methods for solving quadratic equations? What would be the best method to solve #-3x^2+12x+1=0#? How do you solve #-4x^2+4x=9#? What are the two numbers if the product of two consecutive integers is 72? Which method do you use to solve the quadratic equation #81x^2+1=0#? How do you solve #-4x^2+4000x=0#? How do you solve for x in #x^2-6x+4=0#? How do you solve #x^2-6x-16=0# by factoring? How do you solve by factoring and using the principle of zero products #x^2 + 7x + 6 = 0#? How do you solve #x^2=2x#? See all questions in Comparing Methods for Solving Quadratics Impact of this question 1456 views around the world You can reuse this answer Creative Commons License