4. Solve the equation $10 x-6+x^{2}=\sqrt{2 x^{2}-2 x+1} \cdot \sqrt{36+(x+4)^{2}}$
Solution. Transform the equation $10 x-6+x^{2}=\sqrt{2 x^{2}-2 x+1} \cdot \sqrt{36+(x+4)^{2}}$

$10 x-6+x^{2}=\sqrt{x^{2}+x^{2}-2 x+1} \cdot \sqrt{36+(x+4)^{2}} \Rightarrow 6 x-6+x^{2}+4 x=\sqrt{x^{2}+(x-1)^{2}} \cdot \sqrt{6^{2}+(x+4)^{2}}$,

This equation can be interpreted as the scalar product of vectors $\vec{a}=(x-1, x), \quad \vec{b}=(6, x+4)$, written in two ways - the sum of the products of corresponding coordinates and the product of the lengths of the vectors times the cosine of the angle between them, which in this case is 1. The cosine is equal to one if the vectors are collinear, i.e., their coordinates are proportional

$$
\frac{x-1}{6}=\frac{x}{x+4} \Rightarrow x^{2}+3 x-4=6 x \Rightarrow x^{2}-3 x-4=0 \Rightarrow x=\left[\begin{array}{l}
-1 \\
4
\end{array}, \text { but } x=-1\right. \text { is an extraneous root. }
$$

Answer: 4.