Example 3 Given the real-coefficient equation $x^{3}+2(k-1) x^{2}+9 x+5(k-1)=0$ has a complex root with modulus $\sqrt{5}$, find the value of $k$, and solve this equation.
By the theorem of complex conjugate roots for real-coefficient equations, we know that the original equation has one real root and two complex roots with a modulus of $\sqrt{5}$, which are conjugate complex numbers. Let these three roots be $a+b \mathrm{i}, a-b \mathrm{i}, c(a, b, c \in \mathbf{R}, b \neq 0)$, then
$$
a^{2}+b^{2}=5 \text {. }
$$

Combining (1), we get $\left\{\begin{array}{l}2 a+c=-2(k-1), \\ a c=2, \\ c=-k+1 .\end{array}\right.$
From (2) and (4), we know $c=1-k, a=\frac{1}{2}(1-k)$, and substituting these into (3), we get $k=-1$ or 3.
To find the solutions of the original equation: When $k=-1$, the solutions of the original equation are $1+2 \mathrm{i}, 1-2 \mathrm{i}, 2$; when $k=3$, the solutions of the original equation are $-1+2 i,-1-2 i,-2$.