(\text{If } m \text{ and } n (m>n) \text{ are the roots of the equation } 7(x+2a)^2 + 3a^2 = 5a (7x+23a)\text{ where } a > 0, \text{ then what is } 3m -n \text{ equal to?})

Expanding: 7(x² + 4ax + 4a²) + 3a² = 35ax + 115a². This gives 7x² + 28ax + 28a² + 3a² - 35ax - 115a² = 0, so 7x² - 7ax - 84a² = 0, or x² - ax - 12a² = 0. Sum of roots = a, product = -12a². If roots are m and n (m > n): m + n = a, mn = -12a². Then 3m - n = 2m + (m - n) = 2m + √((m+n)² - 4mn) = 2m + √(a² + 48a²) = 2m + 7a. Since m > n and mn < 0, one root is positive, one negative. Solving: m = 4a, n = -3a. So 3m - n = 12a - (-3a) = 15a.