(a-7)(a+3)>(a+1)(a-5)

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Let's first expand both sides of the inequality:

(a-7)(a+3) = a^2 - 4a - 21
(a+1)(a-5) = a^2 - 4a - 5

Now we have:
a^2 - 4a - 21 > a^2 - 4a - 5

Subtracting a^2 and -4a from both sides gives:
-21 > -5

Since -21 is less than -5, this inequality is true for all real numbers a. Therefore, the solution to the inequality (a-7)(a+3) > (a+1)(a-5) is all real numbers a.