In $\triangle PQR$, we have $PQ = QR = 34$ and $PR = 32$.  Find the length of median $\overline{QM}$.
Because $\triangle PQR$ is isosceles with $PQ=QR$, median $\overline{QM}$ is also an altitude:


[asy]
size(100);
pair P,Q,R,M,NN;
P = (0,0);
Q = (0.5,0.9);
R = (1,0);
NN = (0.5,0);
M = (Q+R)/2;
draw(rightanglemark(Q,NN,P,2.5));
draw(P--Q--R--P);
draw(Q--NN);
label("$P$",P,SW);
label("$R$",R,SE);
label("$Q$",Q,N);
label("$M$",NN,S);
[/asy]

We have $MP = PR/2 = 16$, so right triangle $PQM$ gives us  \begin{align*}
QM &= \sqrt{PQ^2 - PM^2}\\
&= \sqrt{34^2 - 16^2}\\
&= \sqrt{(34-16)(34+16)}\\
& = \boxed{30}.
\end{align*}  (We might also have recognized that $PM/PQ = 8/17$, so $QM/PQ = 15/17$.)