Suppose we have a strategy that requires dropping the glass balls a maximum of $N$ times. On the first drop we can start on floor $N$, since if it breaks we can simply take the second ball and try every floor underneath $N$ until the ball breaks, starting from floor 1, then 2, and so on up to the floor $N-1$. $$ $$ If the first ball does not break, then we need to drop the ball $N-1$ floors higher, such that if it does break we can check every floor in-between $N$ and $2N-1$ by dropping the second ball a maximum of $N-2$ more times. $$ $$ If the first glass ball does not break after dropping it from the floor $2N-1$, then using the identical argument as before we need to drop the ball $N-2$ floors higher. Following this logic, the maximum height of the building, $H$, two glass balls can cover is $$ H = N + (N-1) + (N-2) + (N-3) + ... + 1. $$ This is equivalent to $$ H = \sum_{k=1}^N k = \frac{N(N+1)}{2}. $$ All that remains is to let $H = 100$ and solve the quadratic polynomial for $N$. Solving gives $$ N = 14. $$ We need to drop the glass balls a maximum of 14 times to determine $x$.