If we take ceil and floor of $\sqrt{x(x+1)}$ (when it exists) we get $x$ and $x+1$ respectively. Is there an analog of this assuming roots exist in modular arithmetic (at least modulo primes)?

Assuming $\sqrt{x(x+1)}=b$, I know we can solve by solving the quadratic equation $x^2+x-b^2=0$. I just want to know if there is a way to solve without going through the quadratic equation.