A new way to solve Quartics (Reposted)

Since I was unable to explain this properly last time,I have reposted.Hope this clears the concept. We see an easy but rather useful radical here. It works over only positives,I will show you how to use this over$(-\infty,\infty)$ $a+b+c=\sqrt{a^2+b^2+c^2+(2ab+2bc+2ca)}=\sqrt{a^2+b^2+c^2+\sqrt{4a^2b^2+4b^2c^2+4c^2a^2+8abc(a+b+c)}}$ put $x=a+b+c$ :$x=\sqrt{a^2+b^2+c^2+\sqrt{4a^2b^2+4b^2c^2+4c^2a^2+8abcx}}$ Let $x=\sqrt{α+\sqrt{β+γx}}$, then $x^4-2αx^2-γx+(α^2-β)=0--(i)$. Comparing it to the original radical, $\begin{cases}α=a^2+b^2+c^2\\\dfrac{β}{4}=a^2 b^2+b^2 c^2+c^2 a^2\\\dfrac{γ^2}{64}=a^2 b^2 c^2 \end{cases}$. By Vieta’s, we have that $a^2, b^2, c^2$ are roots of the cubic: $z^3-αz^2+\dfrac{β}{4} z-\dfrac{γ^2}{64}=0--(ii)$. Just change $(\alpha,\beta,\gamma)=(\dfrac{-\alpha}{2},\dfrac{\alpha^2-4\gamma}{4},-\beta)$,then (i) becomes $P(x)=x^4+\alpha x^2+\beta x+\gamma$, and (ii) becomes $F(z)=z^3+\dfrac{\alpha}{2}z^2+ \dfrac{\alpha^2-4\gamma}{16}z-\dfrac{\beta^2}{64}$ since $x$ is $a+b+c$, and the equation in $z$ has roots $a^2,b^2,c^2$, we can say that the sum of the square roots of roots of F is a root of P, but wait!, we are squaring $abc$ to get $\beta$ in P. but if $\beta$ is positive, it would give $abc$ negative, which would change things. so we put $x= -x$, and get a negative $\beta$. This would simply result in $-1$ times the sum of the square roots of roots of F.