The floor of $\sqrt{2x}+1/2$ is the ceiling of $(\sqrt{1+8x}-1)/2$

I've been working on this for a while now and I can't figure out how to prove it:$$\left\lfloor \sqrt{2x} + \frac{1}{2}\right\rfloor = \left\lceil \frac{\sqrt{1+8x}-1}{2}\right\rceil.$$

Here $x$ is an integer.