To derive the quadratic formula, I want to find $(m,n)$ such that $$ax^2+bx+c+p=(mx+n)^2+c$$ Expanding the RHS reveals $$ax^2+bx+c+p=m^2x^2+2mnx+n^2+c.$$ Then, $$a=m^2,b=2mn\implies (m,n)=(\pm\sqrt{a},\pm\frac{b}{2\sqrt{a}})$$ From here, isolating and solving for $x$ would be trivial.
Would this general method be correct? I'm confused as to the significance of $p$ since $p=n^2 \implies n=\pm\sqrt{p}$ here as well.
Perhaps it's been a long day or my algebraic skills have worsened, but I just couldn't seem to work out the quadratic formula with this path of thinking; there would always be some extra term...
I'm aware that dividing the entire original equation by $a$ simplifies things a lot, but that should just be a specific case of this general form.
Thanks in advance!
Your formulation isn't quite right. If you're trying to solve $ax^2+bx+c=0$, you could look for $(m,n,p)$ such that $$ ax^2+bx+c=(mx+n)^2-p.\tag1 $$ Setting the LHS of (1) to zero is then the same as setting the RHS to zero, which yields $$ x=\frac{-n\pm\sqrt p}{m}. $$ To obtain $m, n, p$, you proceed as you've done: Expand the RHS of (1) and match coefficients on powers of $x$.