How do you solve #2y^2 + 34y +132 = 0#? Algebra Quadratic Equations and Functions Comparing Methods for Solving Quadratics 1 Answer Mahek ☮ Apr 9, 2018 #2y^2 + 34y +132 = 0# #=> y^2 + 17y +66 = 0# #=> y^2 + 6y +11y+66 = 0# #=> y(y + 6) +11(y+6) = 0# #=> (y+11)(y+6) = 0# Therefore either #y+11=0# or #y+6=0# thats is #y=-11# or #y=-6# 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 1788 views around the world You can reuse this answer Creative Commons License