How do you solve the quadratic #9y^2+6y-8=0# using any method?

Question

2583 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-13T14:18:30

#y=2/3 or y=-4/3#

#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#