Factor out the leading coefficient:
\(ax^2+bx+c=a\left(x^2+\dfrac bax+\dfrac ca\right)\)
Now complete the square:
\(x^2+\dfrac bax+\dfrac ca=x^2+\dfrac bax+\dfrac{b^2}{4a^2}-\dfrac{b^2}{4a^2}+\dfrac ca=\left(x+\dfrac b{2a}\right)^2-\dfrac{b^2}{4a^2}+\dfrac{4ac}{4a^2}=\left(x+\dfrac b{2a}\right)^2-\dfrac{b^2-4ac}{4a^2}\)
Now, redistributing the leading \(a\) gives
\(a\left(x+\dfrac b{2a}\right)^2-\dfrac{b^2-4ac}{4a}\)
