Three, let $[x]$ denote the greatest integer not exceeding the real number $x$. Solve the equation $x^{2}-2[x]-5=0$.


F is, $\left\{\begin{array}{l}x^{2}-2 x-5>-2, \\ x^{2}-2 x-5 \leqslant 0 .\end{array}\right.$ Therefore, $\left\{\begin{array}{l}x>3, \\ 1-\sqrt{6}<x<1+\sqrt{6} .\end{array}\right.$ Hence, $[x]=-2$ or 3.

When $[x]=-2$, the original equation is $x^{2}=1$, yielding $x= \pm 1$. Since $[ \pm 1]= \pm 1$, this contradicts $[x]=-2$, so it is discarded.

When $[x]=3$, the original equation is $x^{2}=11$, yielding $x= \pm \sqrt{11}$. For $[-\sqrt{11}]=-4$, this contradicts $[x]=3$, so it is discarded. When $x=\sqrt{11}$, it meets the conditions. Therefore, the solution to the original equation is $x=\sqrt{11}$.