6) For how many distinct values of the real number $b$ does the equation

$$
x^{2}+b x-16=0,
$$

have two real solutions (possibly coincident) and these are both integers?
(A) Two,
(B) three, (C) four,
(D) five
(E) six.
6. The answer is $(\mathbf{D})$.

Given a quadratic equation that admits two real solutions and has the coefficient of the quadratic term equal to 1, the sum of the solutions coincides with the first-degree term changed in sign, and their product is equal to the constant term. Therefore, if the equation in the problem admits two real solutions, their sum must be equal to $-b$ and their product must be -16. If we further require that the solutions be integers, there are only the following possibilities:

- one of the solutions is 1 and the other is $-16 \Rightarrow b=15$;
- one of the solutions is -1 and the other is $16 \Rightarrow b=-15$;
- one of the solutions is 8 and the other is $-2 \Rightarrow b=-6$;
- one of the solutions is -8 and the other is $2 \Rightarrow b=6$;
- one of the solutions is -4 and the other is $4 \Rightarrow b=0$.

Therefore, $b$ can take 5 distinct values.