When `a != 0`, there are two solutions to `ax^2 + bx + c = 0` and they are

`x = (-b +- sqrt(b^2-4ac))/(2a) .`

When $a \ne 0$, there are two solutions to \(ax^2 + bx + c = 0\) and they are $$x = {-b \pm \sqrt{b^2-4ac} \over 2a}.$$

When a0 , there are two solutions to ax2 + bx + c = 0 and they are x = b ± b2 4ac 2a .