Let the Domain be Set of All Real Numbers.
Then, We can write the Expression for " Every positive real number has exactly two square roots." as following :
1. $\forall x [(x >0) \rightarrow \exists y \exists z((y \neq z) \wedge (y^2 = x) \wedge (z^2 = x) \wedge \forall p((p \neq y,z) \rightarrow (p^2 \neq x)))]$
NOTE that $\forall x [(x >0) \rightarrow \exists y \exists z((y \neq z) \wedge (y^2 = x) \wedge (z^2 = x)]$ make sure that there are at least Two square roots for every positive real number and $\forall p((p \neq y,z) \rightarrow (p^2 \neq x)))]$ makes sure that there No more than Two.... So, There are Exactly Two square roots of every Positive real number.
2. $\forall x((x > 0) \rightarrow \exists y \exists z((y \neq z) \wedge \forall p((x = p^2) \leftrightarrow (p = y \vee p = z))))$
NOTE how this uses a "there exist exactly 2" quantifier :
$\exists ^{=2} x P(x) $ $\equiv$ $\exists y \exists z((y \neq z) \wedge \forall w((P(w)) \leftrightarrow (w = y \vee w = z)))$
(Note that $ \exists ^{=2} $ is the notation here, which essentially is a "two-point generalization" of the "there is exactly one" quantifier :
$\exists ! x P(x)$ $\equiv $ $\exists y \forall z(P(z) \leftrightarrow z = y)$
