Let $a,b,c$ be real numbers, with $a\ne0$. Find the flaw in the proof that $-b/2a$ is a solution of $ax^2+bx+c=0$
(Fake proof.) We have $x$ and $y$ as solutions. According to Viète, $x+y= -b/a$. Since $x$ and $y$ can be any solutions, let $x = y$ we have $2x = -b/a$ or $x=-b/2a$
My attempt is plugging $x = -b/2a$ to verify that $xy = c/a$. If so then $b^2/4a^2 = c/a$, which shows that $ b^2 - 4ac$ must always equals to $0$, which is not true.
I would like a verification for my proof.
Thank you.