\( \Large 3x^{2} + 8x +4 = 0 \)

= \( \Large 3x^{2}+6x+2x+4=0 \)

= \( \Large 3x \left(x+2\right) + 2 \left(x+2\right) = 0 \)

= \( \Large \left(x+2\right) \left(3x+2\right) = 0 \)

Therefore, x = -2, \( \Large -\frac{2}{3} \)

and \( \Large 4y^{2} - 19y + 12 = 0 \)

=> \( \Large 4y^{2} - 16y -3y + 12 = 0 \)

=> \( \Large 4y \left(y - 4\right) - 3 \left(y - 4\right) = 0 \)

=> \( \Large \left(y - 4\right) \left(4y - 3\right) = 0 \)

Therefore, \( \Large y = 4, \frac{3}{4} \)

Hence, y > x or x < y