jmc

Let $r=|a|=|b|=|c|.$ We can re-arrange $a{z}^2+bz+c=0$ as $$a{z}^2=-bz-c.$$By the Triangle Inequality, $$|a{z}^2|=|-bz-c|\le |bz|+|c|,$$so $|a||z{|}^2\le |b||z|+|c|,$ or $r|z{|}^2\le r|z|+r.$ Then $$|z{|}^2\le |z|+1,$$so $|z{|}^2-|z|-1\le 0.$ This factors as $$\left(|z|-\frac{1-\sqrt{5}}{2}\right)\left(|z|-\frac{1+\sqrt{5}}{2}\right)\le 0,$$so $|z|\le \frac{1+\sqrt{5}}{2}.$

The numbers $a=1,$ $b=-1,$ $c=-1,$ and $z=\frac{1+\sqrt{5}}{2}$ satisfy the given conditions, so the largest possible value of $|z|$ is $\boxed{\frac{1+\sqrt{5}}{2}}.$