Question:
Prove that the quadratic congruence 6x^2+5x+1=0(mod p) has a solution for every prime p, even though the equation 6x^2+5x+1=0 has no solution in the integers.
There is a solution x=1 if p=2 or p=3.
The solution in the reals is $\displaystyle x = \frac{-5\pm1}{12}$. If p is not equal to 2 or 3 then 12 is coprime to p, and so 12 will have an inverse $\displaystyle k$ (mod p). Then $\displaystyle x = k(-5\pm1)$ will be solutions to $\displaystyle 6x^2+5x+1=0$ (mod p).