information:filtering, information:extraction, structural:expansion, linguistic:paraphrasing, structural:reformatting
By substitution

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

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

a (u − b 2 a) 2 + b (u − b 2 a) + c = 0 a (u 2 − b a u + b 2 4 a 2) + b (u − b 2 a) + c = 0 a u 2 − b u + b 2 4 a + b u − b 2 2 a + c = 0 a u 2 + 4 a c − b 2 4 a = 0 u 2 = b 2 − 4 a c 4 a 2. {\displaystyle {\begin{aligned}a\left(u-{\frac {b}{2a}}\right)^{2}+b\left(u-{\frac {b}{2a}}\right)+c&=0\\a\left(u^{2}-{\frac {b}{a}}u+{\frac {b^{2}}{4a^{2}}}\right)+b\left(u-{\frac {b}{2a}}\right)+c&=0\\au^{2}-bu+{\frac {b^{2}}{4a}}+bu-{\frac {b^{2}}{2a}}+c&=0\\au^{2}+{\frac {4ac-b^{2}}{4a}}&=0\\u^{2}&={\frac {b^{2}-4ac}{4a^{2}}}.\end{aligned}}}

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

x = − b ± b 2 − 4 a c 2 a. {\displaystyle x={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}.}