6.066. $x^{2}-4 x-6=\sqrt{2 x^{2}-8 x+12}$.
Solution.

By squaring both sides of the equation, we have

\[
\left(x^{2}-4 x-6\right)^{2}=2 x^{2}-8 x+12 \Leftrightarrow\left(x^{2}-4 x-6\right)^{2}-2\left(x^{2}-4 x-6+12\right)=0
\]

Let \(x^{2}-4 x-6=y, y \geq 0\). The equation in terms of \(y\) becomes \(y^{2}-2 y-24=0\), from which \(y_{1}=-4, y_{2}=6 ; y_{1}=-4\) is not valid.

Then \(x^{2}-4 x-6=6 \Leftrightarrow x^{2}-4 x-12=0, x_{1}=-2, x_{2}=6\). By verification, we confirm that these are indeed the roots of the original equation.

Answer: \(x_{1}=-2, x_{2}=6\).

Solve the systems of equations (6.067-6.119):