Chapter preface.
When a≠0a \ne 0a≠0, there are two solutions to (ax2+bx+c=0)(ax^2 + bx + c = 0)(ax2+bx+c=0) and they are x=−b±b2−4ac2a.x = {-b \pm \sqrt{b^2-4ac} \over 2a}.x=2a−b±√b2−4ac.