7.100. $2 \lg x-\lg 4=-\lg \left(5-x^{2}\right)$.
## Solution.

Domain of definition: $\left\{\begin{array}{l}x>0, \\ 5-x^{2}>0,\end{array} \quad 0<x<\sqrt{5}\right.$.

From the condition

$$
\begin{aligned}
& \lg x^{2}+\lg \left(5-x^{2}\right)=\lg 4, \lg \left(x^{2}\left(5-x^{2}\right)\right)=\lg 4, x^{2}\left(5-x^{2}\right)=4 \\
& x^{4}-5 x^{2}+4=0
\end{aligned}
$$

Solving this equation as a biquadratic equation in terms of $x$, we find $x_{1}=-1, x_{2}=1, x_{3}=-2, x_{4}=2 ; x_{1}=-1$ and $x_{3}=-2$ do not satisfy the domain of definition.

Answer: $1 ; 2$.