Question 444397
To find the solutions of quadratic equation {{{ax^2+bx+c=0}}} where a≠0, we

divide both sides by a:{{{(x^2+(b/a)x +c/a)=0}}}, complete the square:

{{{(x^2+(b/a)x+b^2/4a^2)=b^2/4a^2-c/a}}} => {{{(x+b/2a)^2=(b^2-4ac)/4a^2}}}

Based on square root definition we have:{{{x=(-b +- sqrt(b^2-4ac))/2a}}}

If {{{b^2-4ac=0}}} we have one solution:{{{x=-b/2a}}}.

If {{{b^2-4ac}}}≠0 we have two different solutions:

{{{x[1]=(-b+sqrt(b^2-4ac))/2a}}} and {{{x[2]=(-b-sqrt(b^2-4ac))/2a}}}

If {{{b^2-4ac<0}}} the equation has no real solutions.