How do you solve the quadratic #9y^2+6y-8=0# using any method? Precalculus Linear and Quadratic Functions Completing the Square 1 Answer Barney V. Feb 13, 2017 #y=2/3 or y=-4/3# Explanation: #9y^2+6y-8=0# #:.(3y-2)(3y+4)# #:.3y-2=0 or 3y+4=0# #:.3y=2 or 3y=-4# #:.y=3/2 or y=-4/3# substitute #y = 3/2# #:.9(2/3)^2+6(2/3)-8=0# #:.9(4/9)+12/3-8=0# #:.4+4-8=0# #:.8-8=0# substitute #y =-4/3# #:.9(-4/3)^2+6(-4/3)-8=0# #:9(16/9)-24/3-8=0# #:.16-8-8=0# #:.8-8=0# 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 2498 views around the world You can reuse this answer Creative Commons License