x2-2(4a-1)x+15a2>2a+7,
x2-2(4a-1)x>-15a2+2a+7,
x2-2(4a-1)x+(4a-1)2>(4a-1)2-15a2+2a+7, adding (4a-1)2 to complete the square,
(x-4a+1)2>16a2-8a+1-15a2+2a+7,
(x-4a+1)2>a2-6a+8,
(x-4a+1)2>(a-2)(a-4).
Note that the left-hand side is a perfect square so it's positive for all x∈ℝ(it can't be negative).
Therefore, since all positive numbers are greater than all negative numbers, we can see that when the right-hand side is negative, this inequality will always be true for all real x. When 2<a<4, the right-hand side is always negative. For example, if a=3, the right-hand side=-1. So the solution for a is 2<a<4.