6.019. $\frac{21}{x^{2}-4 x+10}-x^{2}+4 x=6$.
## Solution.

Domain: $x \in R$.

From the condition, we have $\frac{21}{x^{2}-4 x+10}-\left(x^{2}-4 x+10\right)+4=0$.

Let $x^{2}-4 x+10=y \neq 0: \frac{21}{y}-y+4=0 \Leftrightarrow \frac{y^{2}-4 y-21}{y}=0$. The equation $y^{2}-4 y-21=0$ has roots $y_{1}=-3, y_{2}=7$. Relative to $x$, we get two equations: $x^{2}-4 x+10=-3, x^{2}-4 x+13=0(D<0)$ or $x^{2}-4 x+10=7, x^{2}-4 x+3=0, x_{1}=3, x_{2}=1$.

Answer: $x_{1}=3, x_{2}=1$.