If the expression √(6 + √(6 + √(6 + …))) continues forever, its value is:

Call the value $x$. Because the pattern inside the first root is the same endless expression, $x$ = √$(6+x)$. Squaring gives $x^2=x+6$, so $x^2-x-6=0$ and $(x-3)(x+2)=0$. A square root cannot be negative, so $x=3$.