Question 1107972
ax^2+bx+c=0
ax^2+bx=-c
x^2+(b/a)x=-c/a, dividing everything by a
complete the square
(x+(b/2a))^2=-(c/a)+b^2/4a^2, adding the (b/2a)^2 to both sides
(x+(b/2a))^2=(-4ac+b^2)/4a^2, putting everything over a common denominator.
x=-(b/2a) +/- sqrt [(b^2-4ac)/4a^2]
x=(-b/2a)+/- sqrt[(b^2-4ac)]/ 2a
x=(1/2a)(-b+/- sqrt (b^2-4ac))