Example Given $91=(6(1)+1)(6(2)+1)=6^2(2)+6(3)+1$. If we wish to find a possible replacement for six we might try solving $2x^2+3x-90=0$. The quadratic equation gives $$x=\frac{-3 \pm 27}{4}$$
Is it generally true that when we arrive at a quadratic equation in this way from a known factorization of a given natural number that it will not have two integer solutions?
Yes. Let all variables be natural numbers (positive integers). $ax^{2}+bx+c=0$. Dividing this equation by $a$ yields $x^2+p*x-c=0$, with $p=\frac{b}{a}$ and $q=\frac{c}{a}$. The solution formula for quadratic equations yields $x=-\frac{p}{2}+\frac{\sqrt(p^{2}+4p)}{2},-\frac{p}{2}-\frac{\sqrt(p^{2}+4p)}{2}$. The second solution is lower or equal to $0$, that is not positive, and so not a natural number.