Solve the following equation in the set of real numbers:

$$
(\sqrt{5}+2)^{x}+(\sqrt{5}-2)^{x}=18
$$
To solve the equation, it is necessary to observe that the two terms on the left side are reciprocals of each other. This is true because

$$
(\sqrt{5}+2) \cdot(\sqrt{5}-2)=(\sqrt{5})^{2}-2^{2}=1
$$

Let $y=(\sqrt{5}+2)^{x}$. Based on the above, $(\sqrt{5}-2)^{x}=\frac{1}{y}$. Substituting these into the equation:

$$
y+\frac{1}{y}=18
$$

Multiplying by $y$ and rearranging, we get the quadratic equation $y^{2}-18 y+1=0$, which has two solutions: $9+\sqrt{80}$ and $9-\sqrt{80}$, which are also reciprocals of each other.

It is easily verifiable that

$$
9+\sqrt{80}=(\sqrt{5}+2)^{2} \quad \text { and } \quad 9-\sqrt{80}=(\sqrt{5}-2)^{2}=(\sqrt{5}+2)^{-2}
$$

Since neither base is 1, it follows from these that the equation is satisfied only by $x=2$ and $x=-2$. The reversibility of our steps also confirms that these are indeed solutions.

Thus, the equation has two solutions: $x_{1}=2$ and $x_{2}=-2$.