9. (16 points) Given non-zero real numbers $a, b, c, t$ satisfy
$$
\left\{\begin{array}{l}
a=t b+c, \\
b=c\left(1+t+t^{2}\right) .
\end{array}\right.
$$
(1) Prove: The quadratic equation
$$
x^{2}+c(b-2 c) x-\left(b^{2}+c^{2}\right)(b-c)=0
$$

must have real roots;
(2) When $a=15, b=7$, find $c, t$.
(1) From $\left\{\begin{array}{l}a=t b+c, \\ b=c\left(1+t+t^{2}\right),\end{array}\right.$ eliminating $t$ yields $b=c\left[1+\frac{a-c}{b}+\left(\frac{a-c}{b}\right)^{2}\right]$.
Rearranging and simplifying gives
$$
c a^{2}+c(b-2 c) a-\left(b^{2}+c^{2}\right)(b-c)=0 .
$$

Therefore, the equation
$$
c x^{2}+c(b-2 c) x-\left(b^{2}+c^{2}\right)(b-c)=0
$$

has a real root $a$.
(2) Substituting $a=15 ; b=7$ into equation (1) and simplifying yields
$$
\begin{array}{l}
c^{3}-37 c^{2}+379 c-343=0 \\
\Rightarrow(c-1)\left(c^{2}-36 c+343\right)=0 .
\end{array}
$$

And $c^{2}-36 c+343=(c-18)^{2}+19>0$.
Thus, $c=1$.
Then, from $a=t b+c$, we get $t=2$.