Following this link, to solve a quadratic equation of the form $x^2+bx+c=0$, We try to factor the equation into $(x-m)(x-n)=x^2-nx-mx+nm$. So we need to have $b=-(n+m)$ and $c=nm$.
To solve for $n,m$, we notice that $n=\frac{-b}{2}+z, m=\frac{-b}{2}-z$ for some $z$. Hence, $c=nm=\frac{b^2}{4}-z^2$. So we solve $z =\pm \sqrt{\frac{b^2}{4}-c}$. So $z$ has 2 possible values, so each of them give n,m. So then there are 4 solutions to quadratic equations?
The second term in $n$ has to be the negative of the second term in $m$. So there are only two solutions.