The value of $\sqrt{6+\sqrt{6+\sqrt{6+\mathrm{...}\infty }}}$ is __.

Let $x=\sqrt{6+\sqrt{6+\sqrt{6+\mathrm{...}\infty }}}$

Then, $x=\sqrt{6+x}$

$\Rightarrow x^2=6+x$

$\Rightarrow x^2-x-6=0$

$\Rightarrow \left(x-3\right)\left(x+2\right)=0$

$\Rightarrow x=3orx=-2$

But the given expression is positive. So, $x-3.$

Hence the value of the given expression is $3.$