Respuesta :

LammettHash
Factor out the leading coefficient:

[tex]ax^2+bx+c=a\left(x^2+\dfrac bax+\dfrac ca\right)[/tex]

Now complete the square:

[tex]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}[/tex]

Now, redistributing the leading [tex]a[/tex] gives

[tex]a\left(x+\dfrac b{2a}\right)^2-\dfrac{b^2-4ac}{4a}[/tex]

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