5. Given real numbers $a, b, c$, and $a>100$, then the number of integers $x$ that satisfy the inequality $\left|a x^{2}+b x+c\right| \leqslant 50$ is
A. Exactly 1
B. At most two
C. Zero
D. Infinitely many
5. B. Suppose there are three different integers $x_{1}, x_{2}, x_{3}$ satisfying $\left|a x^{2}+b x+c\right|x_{1} \geqslant-\frac{b}{2 a}$,
then $a\left(x_{1}+x_{2}\right)+b \geqslant 2 a x_{1}+a+b>a$.
Therefore, $\left|\left(a x_{2}^{2}+b x_{2}+c\right)-\left(a x_{1}^{2}+b x_{1}+c\right)\right|=\left|\left(x_{2}-x_{1}\right)\left[a\left(x_{1}+x_{2}\right)+b\right]\right| \geqslant a>100$.
On the other hand, $\left|\left(a x_{2}^{2}+b x_{2}+c\right)-\left(a x_{1}^{2}+b x_{1}+c\right)\right| \leqslant\left|a x_{2}^{2}+b x_{2}+c\right|+\left|a x_{1}^{2}+b x_{1}+c\right| \leqslant 50+50=$ 100, contradiction, hence the answer is B.