In each of the following question, two equations are given. You have to solve them and give answer- I. (18x^2 + 18x + 4 = 0) II. (12y^2 + 29y + 14 = 0)

Solving Equation I: 18x² + 18x + 4 = 0. Using quadratic formula: x = (-18 ± √(324-288)) / 36 = (-18 ± √36) / 36 = (-18 ± 6) / 36. So x = -12/36 = -1/3 or x = -24/36 = -2/3. Solving Equation II: 12y² + 29y + 14 = 0. Factoring: 12y² + 8y + 21y + 14 = 0, so 4y(3y+2) + 7(3y+2) = 0, giving (4y+7)(3y+2) = 0. Thus y = -7/4 = -1.75 or y = -2/3 ≈ -0.67. Comparing: x values are -0.33 and -0.67, y values are -1.75 and -0.67. Since x = -2/3 = y, but x also has -1/3 which is greater than both y values, we have x ≥ y.