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.