Editing Quadratic formula (section)

===By substitution===
Another derivation uses a [[change of variables]] to eliminate the linear term. Then the equation takes the form {{tmath|1=\textstyle u^2 = s}} in terms of a new variable {{tmath|u}} and some constant expression {{tmath|s}}, whose roots are then {{tmath|1= u = \pm \sqrt s}}.

By substituting {{tmath|1= x = u - \tfrac{b}{2a} }} into {{tmath|1=\textstyle ax^2 + bx + c = 0}}, expanding the products and combining like terms, and then solving for {{tmath|\textstyle u^2\!}}, we have:

<math display=block>\begin{align}
a\left(u-\frac{b}{2a}\right)^2 + b\left(u-\frac{b}{2a}\right) + c &=0 \\[5mu]
a\left(u^2-\frac{b}{a}u+\frac{b^2}{4a^2}\right) + b\left(u-\frac{b}{2a}\right) + c &= 0 \\[5mu]
au^2 - bu + \frac{b^2}{4a} + bu - \frac{b^2}{2a}+c &= 0 \\[5mu]
au^2 + \frac{4ac - b^2}{4a} &= 0 \\[5mu]
u^2 &= \frac{b^2 - 4ac}{4a^2}.
\end{align}</math>

Finally, after taking a square root of both sides and substituting the resulting expression for {{tmath|u}} back into {{tmath|1= x = u - \tfrac{b}{2a},}} the familiar quadratic formula emerges:

<math display=block>
x = \frac{-b\pm \sqrt{b^2 - 4ac}}{2a}.
</math>