False proof that $\tfrac{0}{0} = 2.5$

Question

I was just playing around with the quadratic formula and wondered what if the a coefficient is $0$, so I took a normal equation like $2x-5=0$ and slapped a $0x^2$ term in front of it to get a quadratic equation. So: $0x^2+2x-5=0$ then using the quadratic formula we get that $x=0/0$ & $x=-4/0$ if we solved the first equation the "normal" way we would get that $x = 2.5$, so would that mean that $0/0 = 2.5$ & $-4/0 = 2.5$? I know that the quadratic equation is not defined for $a = 0$, but could you logically prove that this is false? to be more broad, yes you could have it be equal to anything if you engineer the initial equation, and having $0/0$ equal to any number will pretty much break all of mathematics. But besides this is there anything more "logical", and not criticizing that this is false because it is not useful. and i don't even know how to even start to prove that this is wrong. (note: sorry if the formatting looks weird, i don't know how to do latex stuff)

Answer 1

The quadratic equation works only when $a\ne0$. So your method of proof is fine except the first premise when you assumed that the quadratic equation would satisfy a linear equation. However, we can say that the solution to the linear equation $bx+c=0$ is $$\lim_{a\rightarrow0}\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\lim_{a\rightarrow0}\mp\frac{4c}{4\sqrt{b^2-4ac}}=-\frac cb$$Which is true.