Since $abc$ is the decimal representation of a three-digit prime number $p$, we have $a \neq 0$, $c \neq 0$. Suppose, contrary to our claim, that there exists a rational root of the equation $ax^2+bx+c=0$. Then the discriminant of this equation is a perfect square, i.e. $b^2-4ac=n^2$, where $n$ is a positive integer number, $n < b$. Multiplying the equality $p=100a+10b+c$ by $4a$ we have
$$4np = 400a^2 + 40ub + 4ac = 400a^2 + 40ub + b^2 - n^2.$$
Thus $4np = (20a+b)^2 - n^2 = (20a+b+n)(20a+b-n)$. The numbers $20a+b+n$ and $20a+b-n$ have the same parity and since their product is divisible by 4 both of $10a+(b+n)/2$ are positive integers. Since their product is divisible by the prime number $10a+(b+n)/2 \le 10a+b \le 99 < p$. This contradiction proves the required statement.