Simplifying $\frac{-8+\sqrt{48}}{8}$ to $\frac{-2+\sqrt{3}}{2}$

Question

I cannot understand the logic of getting from "unsimplified" to "simplified":

An unsimplified answer: $x=\dfrac{-8+\sqrt{48}}{8}$, $\dfrac{-8-\sqrt{48}}{8}$

A simplified answer: $x=\dfrac{-2+\sqrt{3}}{2}$, $\dfrac{-2-\sqrt{3}}{2}$

(original problem image)

Would someone please explain? Thank you.

Answer 1

$$\frac{-8\pm\sqrt{48}}{8}=\frac{-8\pm\sqrt{16}\cdot\sqrt{3}}{8}=\frac{-8\pm4\sqrt{3}}{8}=\frac{4(-2\pm\sqrt{3})}{4\cdot 2}=\frac{-2\pm\sqrt{3}}{2}$$

Simplification