Following is a proposition from Artin's Algebra
$(3.3)$ Proposition. Assume that the field $F$ does not have characteristic $2$ , that is, that $1+1 \neq 0$ in $F$. Then any extension $F \subset K$ of degree $2$ can be obtained by adjoining a square root: $K=F(\delta)$, where $\delta^2=D$ is an element of $F$. Conversely, if $\delta$ is an element of an extension of $F$, and if $\delta^2 \in F$ but $\delta \notin F$, then $F(\delta)$ is a quadratic extension.
Proof. We first show that every quadratic extension is obtained by adjoining a root of a quadratic polynomial $f(x) \in F[x]$. To do this, we choose any element $\alpha$ of $K$ which is not in $F$. Then $(1, \alpha)$ is a linearly independent set over $F$. Since $K$ has dimension $2$ as a vector space over $F,(1, \alpha)$ is a basis for $K$ over $F$, and $K=F[\alpha]$. It follows that $\alpha^2$ is a linear combination of $(1, \alpha)$, say $\alpha^2=-b \alpha-c$, with $b, c \in F$. Then $\alpha$ is a root of $f(x)=x^2+b x+c$.
Since $2 \neq 0$ in $F$, we can use the quadratic formula $\alpha=\frac{1}{2}\left(-b+\sqrt{b^2-4 c}\right)$ to solve the equation $x^2+b x+c=0$. This is proved by direct calculation. There are two choices for the square root, one of which gives our chosen root $\alpha$. Let $\delta$ denote that choice: $\delta=\sqrt{b^2-4 c}=2 \alpha+b$. Then $\delta$ is in $K$, and it also generates $K$ over $F$. Its square is the discriminant $b^2-4 c$, which is in $F$. The last assertion of the proposition is clear. $\square$
My doubt is the following:
Given any $\alpha \in K$ such that $\alpha \notin F$ I can prove that $\alpha^{2} \notin F$.
$Proof:$ Let $\alpha^{2} \in F$, but since $\alpha^2$ is a linear combination of $(1, \alpha)$, say $\alpha^2=-b \alpha-c$, with $b, c \in F$ it implies
$$-b \alpha-c=d=\alpha^2 \in F$$
$$\implies \alpha=\frac{d+c}{b}\in F$$
which is a contradiction so given any $\alpha \in K$ such that $\alpha \notin F \implies$ $\alpha^{2} \notin F$.
But in the proof in Artin we do find such a $\delta \in K$ such that $\delta ^2 \in F$ but $\delta \notin F$
Please point out where I am making the mistake
If the square is in the base field, then we have $\alpha^2=-b\alpha-c$ with $b=0$,
and you may not divide by $b=0$, i.e., $\dfrac{d+c}b$ is undefined when $b=0$.