One. (20 points) Given that $a$ and $b$ are integers, the equation $a x^{2} + b x + 2 = 0$ has two distinct negative real roots greater than -1. Find the minimum value of $b$.


Let the equation $a x^{2}+b x+2=0(a \neq 0)$ have two distinct negative real roots $x_{1} 、 x_{2}\left(x_{1}<x_{2}<0\right)$.
\end{array}\right.
$$

Solving, we get $a>0, b>0$.
Since $a$ and $b$ are both integers, it follows that $a$ and $b$ are both positive integers.
Let $y=a x^{2}+b x+2$. Then this parabola opens upwards, intersects the $x$-axis at two different points, and when $x=-1$, $y>0$. Therefore,
$$
\begin{array}{l}
\left\{\begin{array} { l } 
{ \Delta = b ^ { 2 } - 8 a > 0 , } \\
{ a - b + 2 > 0 }
\end{array} \Rightarrow \left\{\begin{array}{l}
8 a \leqslant b^{2}-1, \\
a \geqslant b-1
\end{array}\right. \\
\Rightarrow b^{2}-1 \geqslant 8 b-8 \\
\Rightarrow(b-1)(b-7) \geqslant 0 .
\end{array}
$$

Since $b-1 \geqslant 0$, then $b-7 \geqslant 0 \Rightarrow b \geqslant 7$.
When $b=7$, substituting into equations (1) and (2) gives $a=6$, at which point the roots of the original equation are $-\frac{1}{2},-\frac{2}{3}$.
Therefore, the minimum value of $b$ is 7.