In the following questions two equations I and II are given. You have to solve both the equations and give answer. ( (I) 625x^2 – 95x – 6 = 0 ) ( (II) 40y^2 + 29y – 56 = 0 )

For equation I: 625x² - 95x - 6 = 0. Using quadratic formula: x = (95 ± √(9025 + 15000))/1250 = (95 ± √24025)/1250 = (95 ± 155)/1250. Solutions: x = 250/1250 = 0.2 or x = -60/1250 = -0.048. For equation II: 40y² + 29y - 56 = 0. Solutions: y = (-29 ± √(841 + 8960))/80 = (-29 ± √9801)/80 = (-29 ± 99)/80. Solutions: y = 70/80 = 0.875 or y = -128/80 = -1.6. Since x can be greater than y (0.2 < 0.875 is false, but comparing -0.048 to -1.6: -0.048 > -1.6), and x can be less than y (0.2 < 0.875), the relationship cannot be uniquely determined.