Answer
False. The quadratic formula is developed by applying factoring and completing the square.
Work Step by Step
$ax^2+bx+c=0 \longrightarrow a(x^2+\dfrac{bx}{a})=-c \longrightarrow a\bigg[x^2+\dfrac{bx}{a}+\bigg(\dfrac{b}{2a}\bigg)^2-\bigg(\dfrac{b}{2a}\bigg)^2\bigg]=-c \\
\longrightarrow a\bigg[\bigg(x+\dfrac{b}{2a}\bigg)^2-\bigg(\dfrac{b}{2a}\bigg)^2\bigg]=-c \\
\longrightarrow a\bigg(x+\dfrac{b}{2a}\bigg)^2-\dfrac{ab^2}{4a^2}=-c \\
\longrightarrow a\bigg(x+\dfrac{b}{2a}\bigg)^2-\dfrac{b^2}{4a}=-c \\
\longrightarrow a\bigg(x+\dfrac{b}{2a}\bigg)^2=-c+\dfrac{b^2}{4a} \\
\longrightarrow a\bigg(x+\dfrac{b}{2a}\bigg)^2=\dfrac{b^2-4ac}{4a} \\
\longrightarrow \bigg(x+\dfrac{b}{2a}\bigg)^2=\dfrac{b^2-4ac}{4a^2} \\
\longrightarrow \sqrt{\bigg(x+\dfrac{b}{2a}\bigg)^2}=\sqrt{\dfrac{b^2-4ac}{4a^2}} \\
\longrightarrow x+\dfrac{b}{2a}=\pm\dfrac{\sqrt{b^2-4ac}}{2a} \\
\longrightarrow x=-\dfrac{b}{2a}\pm\dfrac{\sqrt{b^2-4ac}}{2a} \\
\longrightarrow x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}$
