The quadratic equation \[ax^2 + bx + c = 0\] has the solution(s) \[x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\] assuming that \(a\neq 0\).

This result looks quite strange the first time you come across it. Until you have seen a proof of it, it might seem a bit magical that it works at all.

In fact, the proof is not very much more than one tool we use in solving quadratic equations anyway—completing the square.

The statements below can be sorted into a proof of the quadratic formula. You might want to print them out and cut them up to rearrange them.

This shows that the original equation is equivalent to \[\left(x + \frac{b}{2a} \right)^2 - \frac{b^2}{4a^2} + \frac{c}{a} = 0.\]

Since \(a \neq 0\), we can divide by \(a\) to get \[x^2 + \frac{b}{a} x + \frac{c}{a} = 0.\]

We complete the square.

We can rewrite the right-hand side by putting it over a common denominator: \[\left(x + \frac{b}{2a} \right)^2 = \frac{b^2 - 4ac}{4a^2}.\]

Consider \(ax^2 + bx + c = 0\), where \(a \neq 0\).

Get the squared term on one side of the equation: \[\left(x + \frac{b}{2a} \right)^2 = \frac{b^2}{4a^2} - \frac{c}{a}.\]

Subtracting \(\frac{b}{2a}\) from both sides and putting the right-hand side over a common denominator gives \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.\]

We can take the square root of both sides.

Since \(x\) appears only once in the equation, we can rearrange this to solve for \(x\).

Taking account of the possibility of positive and negative square roots, we see \[x + \frac{b}{2a} = \frac{\pm\sqrt{b^2 - 4ac}}{2a}.\]