How do you solve #2y = 3y^2#? Algebra Quadratic Equations and Functions Comparing Methods for Solving Quadratics 1 Answer VNVDVI Apr 1, 2018 #y=0, 3/2# Explanation: Divide both sides by #y:# #(2cancely)/cancely=(3y^((cancel2)1))/cancely# #3y=2# #y=2/3# will be one solution. #y=0# is also a solution. Since both sides originally include an instance of #y, y=0# results in #0=0, #a valid solution. 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 2170 views around the world You can reuse this answer Creative Commons License