Root 30

Let $\sqrt{30+\sqrt{30+\sqrt{30+\sqrt{\cdots}}}}=x$ .

Then, $x=\sqrt{30+x}$ .

So, $x^2-x-30=0$ , which can be factored into $(x-6)(x+5)=0$ .

We take the quadratic's positive root to be our answer (since we want the principal square root), so $x=\boxed{6}$ .

$\begin{aligned} \sqrt { 30+\sqrt { 30+\sqrt { 30+\sqrt { 30+\sqrt { 30+\sqrt { 30+\sqrt { \cdots } } } } } } } & \overset { let }{ = } & x \\ \sqrt { 30+x } & = & x \\ 30+x & = & { x }^{ 2 } \\ { x }^{ 2 }-x-30 & = & 0 \\ (x-5)(x+6) & = & 0 \\ x & = & 5\quad (x>0) \\ \sqrt { 30+\sqrt { 30+\sqrt { 30+\sqrt { 30+\sqrt { 30+\sqrt { 30+\sqrt { \cdots } } } } } } } & = & \boxed { 5 } \end{aligned}$