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

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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 $y =-4/3$

$:.9(-4/3)^2+6(-4/3)-8=0$

$:9(16/9)-24/3-8=0$

$:.16-8-8=0$

$:.8-8=0$

How do you solve the quadratic 9y^2+6y-8=09y^2+6y-8=0 using any method?