In the following question two equations are given. You have to solve both the equations and find out the values of x and y and Give answer - ( (I) x^2 – x - 12 = 0 ) ( (II) 2y^2 + 18y + 36 = 0 )

Equation (I): x² - x - 12 = 0. Factorizing: (x-4)(x+3) = 0, so x = 4 or x = -3. Equation (II): 2y² + 18y + 36 = 0. Divide by 2: y² + 9y + 18 = 0. Factorizing: (y+3)(y+6) = 0, so y = -3 or y = -6. Comparing: x values (4, -3) and y values (-3, -6). The smallest x (-3) equals the largest y (-3). All other comparisons give x > y. Therefore x ≥ y is always true, which is option D.