19. Given real numbers $a, b, c$ form an arithmetic sequence, and $a \geqslant b \geqslant c \geqslant 0$. If the quadratic trinomial $a x^{2}+b x+c$ has only one real root, then the real root is ( ).
(A) $-7-4 \sqrt{3}$
(B) $-2-\sqrt{3}$
(C) 1
(D) $-2+\sqrt{3}$
(E) $-7+4 \sqrt{3}$
19. D.

From the problem, we know
$$
\Delta=b^{2}-4 a c=0 \text {. }
$$

Also, $a, b, c$ form an arithmetic sequence, so $b=\frac{a+c}{2}$.
Substituting the above equation into equation (1) yields
$$
\begin{array}{l}
\frac{(a+c)^{2}}{4}=4 a c \\
\Rightarrow\left(\frac{a}{c}\right)^{2}-14\left(\frac{a}{c}\right)+1=0 \\
\Rightarrow \frac{a}{c}=7 \pm 4 \sqrt{3} .
\end{array}
$$

Given $a \geqslant c$, we have $\frac{a}{c}=7+4 \sqrt{3}$.
Thus, the real root of the quadratic trinomial $a x^{2}+b x+c$ is
$$
\begin{aligned}
x & =-\frac{b}{2 a}=-\frac{2 \sqrt{a c}}{2 a}=-\frac{\sqrt{c}}{\sqrt{a}} \\
& =-\sqrt{\frac{1}{7+4 \sqrt{3}}}=-2+\sqrt{3} .
\end{aligned}
$$