How did mathematicians solve the quadratic formula?

Question

How did they solve for x in ax^2+bx+c and suddenly come up with

-b+or-(b^2-4ac)^1/2/2a=x?

Answer 1

ax^2 + bx + c=0
x^2+(b/a)x = -c/a
x^2+(b/a)x + (b^2)/(4a^2)= -(4ac)/(4a^2) + (b^2)/(4a^2)
(x+(b/(2a)))^2 = (b^2-4ac)/(4a^2)
x+(b/(2a)) = +-[Sqrt(b^2-4ac)]/(2a)
x = -(b/2a) +-[Sqrt(b^2-4ac)]/(2a)

x=[-b+-Sqrt(b^2-4ac)]/(2a)

Answer 2

by completing the square.

ax^2+bx+c=[x^2+(b/a)x+(c/a)]*a
x^2+(b/a)x+(c/a) = (x+(b/2a))^2 -(b/2a)^2+c

then you set that equal to zero and solve

Answer 3

i dont believe that is the quadratic formula... isnt it

x= (-b +/- [square root of {b^2-4(a)(c)}])all over 2(a)

and i have no idea.. .we started working on it.. but im confused