138. Solve the equation: $x^{4}-2 x^{2}-400 x=9999$ (solved elementarily).
138. $\quad x^{4}-2 x^{2}-400 x=9999$

$x^{4}-11 x^{3}+11 x^{3}-121 x^{2}+119 x^{2}-1309 x+909 x-9999=0$ $x^{3}(x-11)+11 x^{2}(x-11)+119 x(x-11)+909(x-11)=0$

$$
(x-11)\left(x^{3}+11 x^{2}+119 x+909\right)=0
$$

from here $x-11=0$ and, therefore, $x_{1}=11$ is a root, which Bhaskara gives. But we also have:

$$
\begin{aligned}
& x^{3}+11 x^{2}+119 x+909=0 \\
& x^{3}+9 x^{2}+2 x^{2}+18 x+101 x+909=0 \\
& x^{2}(x+9)+2 x(x+9)+101(x+9)=0 \\
& (x+9)\left(x^{2}+2 x+101\right)=0
\end{aligned}
$$

108
therefore, $x_{2}=-9$ and $x^{2}+2 x+101=0$, from which we will find the imaginary roots $x_{3}$ and $x_{4}$.