Solve the following equation:

$$
(\sqrt{2+3 \sqrt{3}})^{x}+(\sqrt{2-\sqrt{3}})^{x}=4.25
$$
Notice that $\sqrt{2+\sqrt{3}} \cdot \sqrt{2-\sqrt{3}}=1$, and thus any power of this product is also 1. Therefore, the quantities

$$
y=(\sqrt{2+\sqrt{3}})^{x} \quad \text{and} \quad z=(\sqrt{2-\sqrt{3}})^{x}
$$

satisfy the relationships

$$
y+z=4.25, \quad y z=1
$$

From the relationships between the roots and coefficients of a quadratic equation, we know that if the sum and product of two quantities are given, then these quantities are the roots of a quadratic equation where the coefficient of the quadratic term is 1, the coefficient of the linear term is the negative of the sum, and the constant term is the product of the quantities. In our case, $y$ and $z$ are the roots of the equation

$$
u^{2}-4.25 u+1=0
$$

meaning that either root can be chosen as $y$, and the other will always be $z$. The roots of the quadratic equation are

$$
u_{1,2}=\frac{4.25 \pm \sqrt{4.25^{2}-2^{2}}}{2}=\frac{4.25 \pm 3.75}{2}, \quad u_{1}=4, \quad u_{2}=\frac{1}{4}
$$

From this, the two solutions of the original equation are:

$$
(\sqrt{2+\sqrt{3}})^{x_{1}}=4, \quad x_{1}=\frac{\log 4}{\log (\sqrt{2+\sqrt{3}})}=1.575 \ldots
$$

and similarly

$$
x_{2}=\frac{\log 1 / 4}{\log (\sqrt{2+\sqrt{3}})}=-\frac{\log 4}{\log (\sqrt{2+\sqrt{3}})}=-x_{1}
$$