If $y$ = √(6 + √(6 + √(6 + … ))) where the pattern continues without end, then $y$ equals:

Everything under the first radical after the 6 is the whole expression again, so $y$ = √$(6+y)$ with $y$ > 0. Squaring gives $y^2-y-6=0$, i.e. $(y-3)(y+2)=0$. As $y$ is positive, $y=3$.