Good Sum

\hspace{-16}$Let $\bf{\alpha\;,\beta\;,\gamma\;,\delta}$ be the roots of the equation $\bf{x^4-8x^3+bx^2+cx+16=0}$\\\\\\ Where $\bf{\alpha\;,\beta\;,\gamma\;,\delta>0}$\\\\\\ So Using Vieta,s formula\\\\\\ $\bf{\alpha+\beta+\gamma+\delta = 8}$\\\\\\ $\bf{\left(\alpha+\beta\right).(\gamma+\delta)+\alpha.\beta+\gamma.\delta=b}$\\\\\\ $\bf{\alpha.\beta(\alpha+\beta)+\gamma.\delta(\gamma+\delta)=-c}$\\\\\\ $\bf{\alpha.\beta.\gamma.\delta=16}$\\\\\\ Now Here $\bf{\alpha\;,\beta\;,\gamma\;,\delta>0}$, So Apply $\bf{A.M\geq G.M}$\\\\\\\\

\hspace{-16}$$\bf{\frac{\alpha+\beta+\gamma+\delta}{4}\geq \left(\alpha.\beta.\gamma.\delta\right)^{\frac{1}{4}}}$\\\\\\ and Equality hold When $\bf{\alpha=\beta=\gamma=\delta}$\\\\\\ So from Sum or product of roots, We Get\\\\\\ $\bf{\alpha=\beta=\gamma=\delta=2}$\\\\\\ So $\bf{\left(\alpha+\beta\right).(\gamma+\delta)+\alpha.\beta+\gamma.\delta=b\Leftrightarrow b=24}$\\\\\\ $\bf{\alpha.\beta(\alpha+\beta)+\gamma.\delta(\gamma+\delta)=-c\Leftrightarrow c=-32}$\\\\\\ So $\bf{2bx+c=0\Leftrightarrow 48x-32=0}$\\\\\\ So $\boxed{\boxed{\bf{x=\frac{32}{48}=\frac{2}{3}}}}$