The value of $sqrt{[6+sqrt{{6+}} sqrt{(6+ldots ext { upto } infty)}]}$ is equal to

$x=sqrt{6+sqrt{6+sqrt{6+ldots ldots ldots infty}}}$

On squaring,

$x^{2}=6+sqrt{6+sqrt{6+ldots ldots infty}}$

$Rightarrow x^{2}=6+x$

$Rightarrow x^{2}-x-6=0$

$Rightarrow x^{2}-3 x+2 x-6=0$

$Rightarrow x(x-3)+2(x-3)=0$

$Rightarrow(x-3)(x+2)=0$

$Rightarrow x=3$ because $x eq-2$