Let $n \geq 1$ be an integer. Recall that a number $t$ is triangular if there exists an integer $r \geq 0$ such that $t = \binom{r + 1} 2,$ where we take $\binom 1 2 = 0$ by convention.
It appears to me that the following statement is true.
Claim. If $t = \binom{r + 1} 2$ is the smallest triangular number such that $n \leq t,$ then $$r = \sqrt{2t + \frac 1 4} - \frac 1 2 = {\left \lceil \sqrt{2n + \frac 1 4} - \frac 1 2 \right \rceil}.$$
One can verify the first equality by $t = \dfrac{r(r + 1)} 2$; yet, the second equality stumps me.
I would appreciate any comment or advice. Thank you for your time and consideration.
Let $f(n) = \sqrt{2n + \frac14} -\frac12$.
The equation $r = \sqrt{2t + \frac 1 4} - \frac 1 2$ for $t = \binom{r+1}{2}$ tells us that $f(\binom{r+1}{2}) = r$. Similarly, we know that $f(\binom r2) = r-1$.
Also, $f(n)$ is an increasing function of $n$. Therefore for $\binom r2 < n \le \binom{r+1}{2}$, we have $r-1 < f(n) \le r$. But this is exactly what we need to know to conclude that $\lceil f(n) \rceil = r$.
To recap: if $\binom r2 < n \le \binom{r+1}{2}$ (in other words, if $r$ is the least value such that $n \le \binom{r+1}{2}$), then $\left\lceil\sqrt{2n + \frac14} -\frac12 \right\rceil = r$.