Originally Posted by
NonCommAlg clearly every integer is a solution. so i'll assume that $\displaystyle x \notin \mathbb{Z}.$ let $\displaystyle [x]=n$ and $\displaystyle x=n+\theta,$ where $\displaystyle 0 < \theta < 1.$ then $\displaystyle x$ is a solution if and only if: $\displaystyle 0 \leq \theta^2 + 2(n-1) \theta < 1.$
so $\displaystyle \theta^2 + 2(n-1) \theta = \theta(\theta + 2n - 2) \geq 0.$ hence $\displaystyle 0 \leq \theta + 2n - 2 < 2n - 1.$ thus we must have $\displaystyle n \geq 1.$ the converse is obviously true, i.e. if $\displaystyle n \geq 1,$ then $\displaystyle \theta^2 + 2(n-1) \theta \geq 0.$
now, from quadratic formula it's clear that $\displaystyle \theta^2 + 2(n-1)\theta - 1 < 0,$ if and only if $\displaystyle 0 < \theta < 1 - n + \sqrt{n^2 - 2n + 2}.$ thus $\displaystyle n < x < 1+ \sqrt{n^2 - 2n + 2}.$ so the general solution of
your equation is: $\displaystyle \mathbb{Z} \cup \bigcup_{n=1}^{\infty} (n, 1 + \sqrt{n^2 - 2n + 2}).$
by the way questions of this type have nothing to do with calculus and they should be posted in high school pre-algebra subforum.