260. One of the roots of the biquadratic equation is $2+\sqrt{3}$. Write this equation.
260. 

$$
\begin{gathered}
x_{1}=2+\sqrt{3}, x_{2}=2-v^{3} \\
(x_{3}=-2-\sqrt{3}, x_{4}=-2+\sqrt{3}) \\
x_{1}^{2}=5+4 \sqrt{3}, x_{2}^{2}=5-4 \sqrt{3} \\
p=-10, q=25-48=-23
\end{gathered}
$$

The equation will be $x^{4}-10 x^{2}-23=0$.