One, (20 points) Given that $a$, $b$, and $c$ satisfy the system of equations
$$
\left\{\begin{array}{l}
a+b=8, \\
a b-c^{2}+8 \sqrt{2} c=48 .
\end{array}\right.
$$

Try to find the roots of the equation $b x^{2}+c x-a=0$.
(A Paper) One, obviously $a, b$ are the roots of the equation $x^{2}-8 x+c^{2}-8 \sqrt{2} c$ $+48=0$.
From the discriminant $\Delta \geqslant 0$, we get $c=4 \sqrt{2}$.
Thus, $\left\{\begin{array}{l}a b=16, \\ a+b=8 .\end{array}\right.$
Solving it, we get $a=b=4$.
Therefore, the original quadratic equation becomes $x^{2}+\sqrt{2} x-1=0$.
Solving it, we get $x_{1}=\frac{-\sqrt{2}+\sqrt{6}}{2}, x_{2}=-\frac{\sqrt{2}+\sqrt{6}}{2}$.