A portion of the quadratic formula proof is shown. Fill in the missing reason.

\begin{tabular}{|l|l|}
\hline \multicolumn{1}{|c|}{ Statements } & \multicolumn{1}{|c|}{ Reasons } \\
\hline [tex]$a x^2 + b x + c = 0$[/tex] & Given \\
\hline [tex]$a x^2 + b x = -c$[/tex] & Subtract [tex]$c$[/tex] from both sides of the equation \\
\hline [tex]$x^2 + \frac{b}{a} x = -\frac{c}{a}$[/tex] & Divide both sides of the equation by [tex]$a$[/tex] \\
\hline [tex]$x^2 + \frac{b}{a} x + \left( \frac{b}{2a} \right)^2 = -\frac{c}{a} + \left( \frac{b}{2a} \right)^2$[/tex] & Complete the square and add [tex]$\left( \frac{b}{2a} \right)^2$[/tex] to both sides \\
\hline [tex]$x^2 + \frac{b}{a} x + \left( \frac{b}{2a} \right)^2 = -\frac{c}{a} + \frac{b^2}{4a^2}$[/tex] & Square [tex]$\left( \frac{b}{2a} \right)$[/tex] on the right side of the equation \\
\hline [tex]$x^2 + \frac{b}{a} x + \left( \frac{b}{2a} \right)^2 = -\frac{4ac}{4a^2} + \frac{b^2}{4a^2}$[/tex] & Find a common denominator on the right side of the equation \\
\hline [tex]$x^2 + \frac{b}{a} x + \left( \frac{b}{2a} \right)^2 = \frac{b^2 - 4ac}{4a^2}$[/tex] & Add the fractions together on the right side of the equation \\
\hline
\end{tabular}