Let $K=\mathbb{C}(a_1,a_2)$ and consider the polynomial $p(x)=x^2+a_1x+a_2\in K[x]$. Let $M$ be a splitting field of $p(x)$ and write $M=\mathbb{C}(x_1,x_2)$, where $x_1$ and $x_2$ are the roots of $p$. Then we have $p(x)=x^2-(x_1+x_2)x+x_1x_2$ and thus $a_1=-(x_1+x_2), a_2=x_1x_2$ are precisely (up to a sign) the elementary symmetric polynomials in the variables $x_1,x_2$. In other words, $K=M^{\mathfrak{S}_2}$ is the fixed field by the action of the Galois group $\mathfrak{S}_2$.
Now let us consider a sub-field $K_0\subseteq K$ obtained in the following way: If we consider the change of variable $y=u_0+u_1x$ (with $u_0,u_1\in \mathbb{C}$) we check that $x^2+a_1x+a_2=0$ becomes $y^2+b_1y+b_2=0$, where $b_1=b_1(u_0,u_1)=a_1u_1+2u_0$ and $b_2=b_2(u_0,u_1)=u_0^2-a_1u_0u_1+a_2 u_1^2$. Suppose for simplicity that $b_1=0$ ($\Leftrightarrow u_0 =-\frac{a_1}{2}u_1$) and hence $b_2=u_1^2\left(a_2-\frac{a_1^2}{4}\right)$, and define $K_0$ to be the sub-field of $K$ generated by $b_1$ and $b_2$, i.e., $K_0=\mathbb{C}\left(a_2-\frac{a_1^2}{4}\right)\subseteq \mathbb{C}(a_1,a_2)$.
My question is the following: How to find explicitly (and not merely existence results given by general field theory) a sub-field $M_0$ of $M$ which contains $K_0$ and such that $M_0^{\mathfrak{S}_2}=K_0$ ?
$$K= \Bbb{C}(b^2-4c,b)=\Bbb{C}(b,c), \quad M = \Bbb{C}(\sqrt{b^2-4c},b) \quad \scriptstyle\text{the splitting field of } x^2+bx+c \in K[x]$$ $$K = M^\sigma, \qquad \sigma(\sqrt{b^2-4c}) = -\sqrt{b^2-4c},\qquad \sigma(b)=b$$
And what you asked is the one-dimensional subfield $$M_0 = \Bbb{C}(\sqrt{b^2-4c}), \qquad K_0 = M_0^\sigma = \Bbb{C}(b^2-4c)$$