Let $t=\sqrt{x}$. Then we have $$t^2-7
Solving, we get \begin{align} t^2-t-10&<0\\\ \left(t-\frac12\right)^2-\frac{41}4&<0\\\ \left(t-\frac12\right)^2&<\frac{41}4\\\ \left|t-\frac12\right|&<\frac{\sqrt{41}}2\\\ -\frac{\sqrt{41}}2
Substituting, we get $$\frac12-\frac{\sqrt{41}}2<\sqrt{x}<\frac12+\frac{\sqrt{41}}2\implies0\leq x<\left(\frac12+\frac{\sqrt{41}}{2}\right)^2$$ and we get $\boxed{x\in\left[0,\left(\dfrac{1}{2}+\dfrac{\sqrt{41}}{2}\right)^2\right)}$.