You have the following equation:
(x² - 7x + 11)ˣ²⁻¹¹ˣ⁺³⁰ = 1
In order to determine the integer solutions to the previous equation, consider that any quanity powered to zero is equal to 1.
Then, it is necessary to determine the zeros of the polynomial which is an exponent, that is, it is necessary to find the zeros of x² - 11x + 30.
Thus, you factorize the previous polynomial, by using the quadratic formula, which is given by:
x = (-b ± √(b² - 4ac))/2a
where for this case a = 1, b = -11 and c = 30. Replace this values into the quadratic formula:
x = (-(-11) ± √((-11)² - 4(1)(30)))/2(1)
x = (11 ± √(121 - 120))/2
x = (11 ± 1)/2
Then, you have the following two solutions (remind that for a quadratic polynomial there are two zeros):
x = (11 + 1)/2 = 12/2 = 6
x = (11 - 1)/2 = 10/2 = 5
Hence, for the values x = 6 and x = 5, the exponent of the given equation is zero, and any quanitity powered to zero is 1. Then, this values are the integers solutions.
x = 5
x = 6