Indian mathematician Shridharacharya and Persian Mathematician Omar Khayyam (may be others also) are known to have found the formula for roots of a quadratic equation $ax^2+bx+c=0$ as $$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}~~~~(1)$$ independently by forming perfect square SLIGHTLY differently.
Currently, Dr. Proshen Loh of Carnegie Mellon University
has come up with a SLIGHTLY different alternative for finding the roots by solving $$(-\frac{b}{2a}+u) (-\frac{b}{2a}-u)=\frac{b^2}{4a^2}-u^2=\frac{c}{a}~~~~(2)$$ So $x_1,x_2=(-\frac{b}{2a}\pm u)$ will be the roots whose sum is $-b/a$ and product is $c/a$.
One more SLIGHTLY different idea will be to remove second (linear) term of the quadratic as in the case of a cubic:$ax^3+bx^2+cx+d=0$ by using $x=u+p$, then we get $$a(u+p)^2+b(u+p)+c=0 \implies au^2+(2ap+b)u+ap^2+bp+c=0~~~~(3)$$ By setting $2ap+b=0 \implies p=-\frac{b}{2a}$ in (3), we again get the equation (2) for $u$.
The question is: can there be another SLIGHTLY different way for finding the roots of a quadratic equation.