Second grade formulae
Second grade formulae
When $a \ne 0$, there are two solutions to \(ax^2 + bx + c = 0\) and they are
$$x = {-b \pm \sqrt{b^2-4ac} \over 2a}.$$
So, it is holds
- When \(b^2 = 4ac \Rightarrow \) equation has a unique real solution.
- When \(b^2 > 4ac \Rightarrow \) equation has two different real solutions.
- When \(b^2 < 4ac \Rightarrow \) equation has different complex congugate solutions.
Welcome to Sites mathstools. This is your first post. Edit or delete it, then start blogging!
Hi, this is a comment.
To delete a comment, just log in and view the post's comments. There you will have the option to edit or delete them.