How do you solve `x^2 - 4x +13 = 0`?

To solve this equation, rather than the classic formula, completing the square may come in handy.

Completing the square means that we can try to find a binomial square "hidden" in the equation and isolate it, and then deal with the rest.

The formula for the square of a binomial is the following:

`(a+b)^2 = a^2+2ab+b^2`.

So, we need two squares, and a third terms, which is twice the multiplication of the bases of the squares.

Your equation starts with `x^2-4x`. Of course, `x^2` is the square of `x`, so we wanto `-4x` to be twice the multiplication of `x` and another number. This other number is obviously `-2`, so we can conclude that

`(x-2)^2 = x^2-4x+4`.

Your equation differs this expression for a difference of `9` unit, in fact

`x^2-4x+13= (x^2-4x+4)+9`

From this point, we are able to tell that the equation has no solution: we want

`(x^2-4x+4)+9=(x-2)^2+9`

to be zero, but since a square is always positive, how can a sum of two positive quantities be zero? In other terms, you would have

`(x-2)^2+9=0 iff (x-2)^2=-9`

and again, a square can't be equal to a negative number.