Describing steps in an equation

Somehing like this? (two && before the \mbox)

\documentclass[letterpaper]{article}
\usepackage{fullpage}
\usepackage{amsmath,amssymb,amsthm,enumitem}
\usepackage[dvipsnames]{xcolor}
\newcommand{\red}[1]{%
  {\color{OrangeRed}#1}}

\begin{document}
Let us apply completing the square to the standard form of a quadratic equation, $ax^2+bx+c=0$.

\begin{align*}
  ax^2+bx+c&=0 &&\text{Divide by $a$, the leading coefficient.}&\\[1.25ex]
  x^2+\frac{b}{a}x+\frac{c}{a}&=0 &&\text{Transpose the constant term to the RHS.}&\\[1.25ex]
  x^2+\frac{b}{a}x&=-\frac{c}{a}&&\text{\red{Add $\left(\frac{1}{2}\left(\frac{b}{a}\right)\right)^2$ to both sides.}}&\\[1.25ex]
  x^2+\frac{b}{a}x+\frac{b^2}{4a^2}&=-\frac{c}{a}+\frac{b^2}{4a^2}&&\text{Factor the LHS and simplify the RHS.}&\\[1.25ex]
  \left(x+\frac{b}{a}\right)^2&=\frac{b^2-4ac}{4a^2}&&\text{Take the square root of both sides.}&\\[1.25ex]
  x+\frac{b}{a}&=\pm\frac{\sqrt{b^2-4ac}}{2a}&&\text{Solve for $x$.}&\\[1.25ex]
  x&=\frac{-b\pm\sqrt{b^2-4ac}}{2a}&&\text{\textbf{Quadratic Formula}.}&
\end{align*}
\end{document}

Package witharrows provides a new environment WithArrows similar to align where it's possible to use command \Arrow to print some text beside an arrow going from one line to another one.

Here you have the possible result with OP's examples:

\documentclass[letterpaper]{article}
\usepackage{fullpage}
\usepackage{amsmath,amssymb,amsthm,enumitem}
\usepackage[dvipsnames]{xcolor}
\usepackage{tikz}
\usetikzlibrary{calc, arrows.meta, bending, decorations.pathreplacing}

\usepackage{witharrows}
\newcommand{\red}[1]{%
{\color{OrangeRed}#1}}

\begin{document}
Let us show that by completing the square, we can achieve this.
\[        \begin{WithArrows}
        f(x)&=ax^2+bx+c \Arrow{Factor $a$ from $ax^2+bx$. Do you know why?}\\
            &=a\left(x^2+\frac{b}{a}x\right)+c \Arrow[tikz=OrangeRed]{Complete the square by adding $\frac{b^2}{4a^2}$.} \\
            &=a\left(x^2+\frac{b}{a}x+\red{\frac{b^2}{4a^2}} \right)+c\red{-a\left(\frac{b^2}{4a^2}\right)}\Arrow{Factor and simplify.}\\
            &=a\left(x+\frac{b}{2a}\right)^2+c-\frac{b^2}{4a}\\
            &=a\left(x+\frac{b}{2a}\right)^2+\frac{4ac-b^2}{4a}\\
            &=a(x-h)^2+k
        \end{WithArrows}
\] \end{document}

\documentclass[letterpaper]{article}
\usepackage{fullpage}
\usepackage{amsmath,amssymb,amsthm,enumitem}
\usepackage[dvipsnames]{xcolor}
\newcommand{\red}[1]{%
{\color{OrangeRed}#1}}

\usepackage{tikz}
\usetikzlibrary{calc, arrows.meta, bending, decorations.pathreplacing}

\usepackage{witharrows}

\begin{document}
Let us apply completing the square to the standard form of a quadratic equation, $ax^2+bx+c=0$.

\[\begin{WithArrows}
    ax^2+bx+c&=0 \Arrow{Divide by $a$, the leading coefficient.}\\
    x^2+\frac{b}{a}x+\frac{c}{a}&=0 \Arrow{Transpose the constant term to the RHS.}\\
    x^2+\frac{b}{a}x&=-\frac{c}{a} \Arrow[tikz=OrangeRed]{Add $\left(\frac{1}{2}\left(\frac{b}{a}\right)\right)^2$ to both sides.}\\
    x^2+\frac{b}{a}x+\frac{b^2}{4a^2}&=-\frac{c}{a}+\frac{b^2}{4a^2} \Arrow{Factor the LHS and simplify the RHS.}\\
    \left(x+\frac{b}{a}\right)^2&=\frac{b^2-4ac}{4a^2} \Arrow{Take the square root of both sides.}\\
    x+\frac{b}{a}&=\pm\frac{\sqrt{b^2-4ac}}{2a} \Arrow{Solve for $x$.}\\
    x&=\frac{-b\pm\sqrt{b^2-4ac}}{2a}
    \end{WithArrows}
\] \end{document}

Note: Package documentation doesn't mention it, but we should load TikZ and calc, arrows.meta, bending, decorations.pathreplacing tikzlibraries to use WithArrows environment.