Does the following count as a valid derivation of the axis of symmetry?
${ax}^2+bx+c=0$
$\longrightarrow\ a\left(\frac{ax^2}{a}+\frac{bx}{a}\right)+c=0$
$\longrightarrow\ a\left(x^2+\frac{b}{a}x\right)+c=0$
$\longrightarrow{a\left(x+\frac{b}{2a}\right)}^2-a\left(\frac{b}{2a}\right)^2+c=0$
$\longrightarrow{a\left(x+\frac{b}{2a}\right)}^2=a\left(\frac{b}{2a}\right)^2-c$
$\longrightarrow\left(x+\frac{b}{2a}\right)^2=\frac{a\left(\frac{b}{2a}\right)^2-c}{a}$
$\longrightarrow\left(x+\frac{b}{2a}\right)^2=\frac{a\left(\frac{b}{2a}\right)^2}{a}-\frac{c}{a}$
$\longrightarrow\left(x+\frac{b}{2a}\right)^2=\left(\frac{b}{2a}\right)^2-\frac{c}{a}$
$\longrightarrow\ x+\frac{b}{2a}=\pm\sqrt{\left(\frac{b}{2a}\right)^2-\frac{c}{a}}$
$\longrightarrow\ x=-\frac{b}{2a}\pm\sqrt{\left(\frac{b}{2a}\right)^2-\frac{c}{a}}$
The conjugate pair tells us that the x-coordinates of the parabola lie at $\pm\sqrt{\left(\frac{b}{2a}\right)^2-\frac{c}{a}}$ from $x=-\frac{b}{2a}$. Since $\pm\sqrt{\left(\frac{b}{2a}\right)^2-\frac{c}{a}}$ represents the same amount, just in opposite directions, $x=-\frac{b}{2a}$ must represent the equation of the axis of symmetry.