How do you solve #3y^2+2y-1=0#? Algebra Quadratic Equations and Functions Comparing Methods for Solving Quadratics 1 Answer jk.13 Mar 30, 2018 #y=-1, 1/3# Explanation: #3y^2+3y-y-1=0# find two integers that ADD to 2, MULTIPLY to -3 #(3*-1)# #3y(y+1)-(y+1)# factor out common factor #(3y-1)(y+1)# #3y-1 = 0# OR #y+1=0# #y=1/3# OR #y=-1# 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 4344 views around the world You can reuse this answer Creative Commons License