ax²+bx+c=0
Taking 'a' common,
a{x²+(b/a)x+(c/a)}=0
Since a is not equal to 0, so
x²+(b/a)x+(c/a)=0
{x+(b/2a)}²+(c/a)-(b/2a)²=0
{x+(b/2a)}²+(c/a)-(b²/4a²)=0
{x+(b/2a)}²=(b²/4a²)-(c/a)
{x+(b/2a)}²=(b²-4ac)/4a²
{x+(b/2a)}=±√{(b²-4ac)/4a²}
x=(b/2a)±√{(b²-4ac)/4a²}...