SG. 2 Let $[x]$ be the largest integer not greater than $x$. If $B=[10+\sqrt{10+\sqrt{10+\sqrt{10+\cdots}}}]$, find the value of $B$.
Reference 2007 FG2.2 $\ldots x \geq 3+\sqrt{3+\sqrt{3+\sqrt{3+\sqrt{3+\sqrt{3}}}}} \ldots$
Let $y=\sqrt{10+\sqrt{10+\sqrt{10+\cdots}}}$
$$
\begin{array}{l}
y^{2}=10+\sqrt{10+\sqrt{10+\sqrt{10+\cdots}}}=10+y \\
y^{2}-y-10=0 \\
y=\frac{1+\sqrt{41}}{2} \text { or } \frac{1-\sqrt{41}}{2} \quad \text { (rejected) } \\
6<\sqrt{41}<7 \Rightarrow \frac{7}{2}<\frac{1+\sqrt{41}}{2}<4 \\
13.5<10+\sqrt{10+\sqrt{10+\sqrt{10+\cdots}}}<14 \\
B=13
\end{array}
$$