9y^2 + 12y - 5 = 0

( 3y - 1)(3y + 5) = 0

Hence

3y - 1 = 0

y = 1/3

&

3y + 5 = 0

y = -5/3

(3*x + 5)*(3*x - 1) = 0

x = -5/3 or x = 1/3

9y² + 12y - 5 = 0

Use quadratic formula

y = [-12 ± √(12² – 4·9(-5))] / [2·9]

= [-12 ± √324] / 18

= [-12 ± 18] /18

= -1⅔, ⅓

9y² + 12y - 5 = 9(x+1⅔)(x-⅓) = (3x+5)(3x-1)

(3y-1)(3y+5) = 0