Verified answer

7^(x - 5) > 3^(x² + x - 30)

--------------------

разложим x² + x - 30 = (x + 6)(x - 5)

D=1 + 120 = 121 = 11²

x₁₂ = (-1 +- 11)/2 = -6 и 5

-----------

7^(x - 5) > 3^(x + 6)(x - 5)

прологарифмируем по основанию допустим 7

log(7) 7^(x - 5) > log(7) 3^(x + 6)(x - 5)

(x - 5) -  (x + 6)(x - 5) log(7) 3 > 0

(x - 5) (xlog(7) 3  + 6log(7) 3  - 1)  > 0

корни 5 и (1 - 6log(7) 3)/log(7) 3  < 0

(1 - 6log(7) 3)/log(7) 3 < x < 5