Is there a hypergeometric function form of the quadratic formula that converges for all a,b,c? After reading this question about a general formula for roots, I was curious about the simple case of $N=2$.
As is well known to school children, if $ ax^2 + bx + c = 0 $, then the roots are $ \frac{-b \pm \sqrt{b^2-4ac}}{2a} $
Shuffling the binomial series around yields $$  -\frac{c}{b} \sum_{k=0}^{\infty} \binom{-\frac{1}{2}+k}{k} \left( \frac{4ac}{b^2} \right)^k \frac{1}{(k+1)} = -\frac{b}{2a} +  \frac{ \sqrt{b^2-4ac}}{2a} $$ but this doesn't converge for simple cases such as $ a=1,b=1,c=-1 $ since the radius of convergence is $<\frac{1}{4}$.
Is there a hypergeometric series that converges for all $a,b,c$ (where there are roots of course)?

$$S_n=-\frac{c}{b} \sum_{k=0}^{n} \binom{-\frac{1}{2}+k}{k} \left( \frac{4ac}{b^2} \right)^k \frac{1}{(k+1)} $$
$$S_n=-\frac{b \left(1-\sqrt{1-\frac{4 a c}{b^2}}\right)}{2 a}+\frac{c   }{b
   }\,\frac{\binom{n+\frac{1}{2}}{n+1}}{n+2 }\,\left(\frac{4a c}{b^2}\right)^{n+1}\, _2F_1\left(1,\frac{2n+3}{2};n+3;\frac{4 a c}{b^2}\right)$$ where appears the Gaussian hypergeometric function.
