Using complex numbers, you can have negative squares, since [tex] i^2=-1 [/tex]
So, you can find the solutions to
[tex] x^2+20=0 \iff x^2 = -20 \iff x=\pm\sqrt{-20}=\pm i \sqrt{20}=\pm 4i\sqrt{5} [/tex]
So, using the factorization [tex] (x-x_1)(x-x_2) [/tex] where [tex] x_{1,2} [/tex] are the roots of the quadratic equation, you have
[tex] x^2+20 = (x-4i\sqrt{5})(x+4i\sqrt{5})[/tex]
This expression can't be further factorized, because all terms have degree 1.
