This is from Artin's Algebra (2nd ed.), Proposition 15.3.3:
I think that we showed in the first para of the proof that degree two extension can be obtained by adjoining a square root. Then, why they put 'conversely' almost at the end of the para?
How do we know that $\sqrt D$ indeed exists?
In the first paragraph, they show that a degree-2 extension can be obtained by adjoining a root of some quadratic polynomial (not necessarily a square root). Then in the second paragraph they show how the same extension can be obtained just by adjoining a square root (a square root of $D$).
The two square roots of $D$ in $K$ (and thus the two possible choices for $\delta$) are given by $$ \pm(2\alpha + b) $$ You can square this yourself (and simplify using $\alpha^2 = -b\alpha - c$) and see that we do indeed get $D = b^2 - 4c$ as a result.
Finally, the "conversely" part is about showing that adjoining a square root (which is not already in $F$) always gives a degree-2 extension.