Let f(x)=2x^2+x-6 = (2x-3)(x+2).
Suppose a positive integer a divides both 2x-3 and x+2.
Then a must also divide 2(x+2)-(2x-3)=7.
Hence,a can either be 1 or 7.
As a result,2x-3=7^n or -7^n for some positive integer n.
We consider the following cases:
(2x-3) = 1.Then x = 2,which yields f(x) = 4,a prime power.
(2x-3) = -1.Then x = 1,which yields f(x) = -3,not a prime power.
(2x-3) = 7.Then x = 5,which yields f(x) = 49,a prime power.
(2x-3) = -7.Then x = -2,which yields f(x) = 0,not a prime power.
(2x-3) = 7^n or -7^n,for n >= 2.Then,since x + 2 = ((2x-3) + 7)/2,we have that x + 2 is divisible by 7
but not by 49.Hence x + 2 = 7 or -7,yielding x = 5,-9.The former has already been considered,
while the latter yields f(x) = 147.
So x can be either 2 or 5.