jmc

We do not know that $b$ is rational, so we cannot conclude that the radical conjugate of $3+\sqrt{5},$ or $3-\sqrt{5},$ must also be a root of the equation. Instead, we turn to Vieta's formulas: the sum of the roots of the equation is $3,$ so the other root of the equation must be $3-(3+\sqrt{5})=-\sqrt{5}.$ Then $b$ equals the product of the roots: $$b=-\sqrt{5}(3+\sqrt{5})=\boxed{-3\sqrt{5}-5}.$$Alternatively, since $3+\sqrt{5}$ is a root of the equation, $$(3+\sqrt{5}{)}^2-3(3+\sqrt{5})+b=0.$$Hence, $b=\boxed{-3\sqrt{5}-5}.$