Unified method to find formula for polynomial equation roots.

Here is unified solution routine to find formula for any $nth$ degree equation ,  for $n < 5$
We can write The general form for any polynomial equation of any $nth$ degree as:
$$\sum_{j=0}^{n} a_j x^j =a_0+a_1x+a_2x^2+\cdots + a_nx^n =0$$
where $a_n \neq 0$

After dividing the equation by $a_n$ , the substitution $x=y-\frac{a_{n-1}}{na_n}$ will eliminate the term $a_{n-1}x^{n-1}$ so we have,
$$\sum_{j=0}^{n}\left(b_jy^j\right) -b_{n-1}y^{n-1}=b_0+b_1y+b_2y^2+\cdots+b_{n-2}y^{n-2}+b_ny^n$$
As we know that the quadratic equation is already solved by this substitution , but remain the cubic and the quartic which will be solved in the following steps:

For the cubic equation, I will rewrite it  as,
$$y^3+py+q=0 \tag1$$
Let $y=z_1+z_2$
$y^3=z_1^3+3z_1^2z_2+3z_1z_2^2+z_2^3=z_1^3+3z_1z_2(z_1+z_2)+z_2^3$
$$y^3-3z_1z_2x-(z_1^3+z_2^3)=0 \tag 2$$

By comparing the coefficients of $eq(2)$ with the original equation (1) we get,
$p=-3z_1z_2 \Rightarrow z_1^3z_2^3=\frac{-p^3}{27}$
and
 $q=-(z_1^3+z_2^3)$
assume that $z_1^3$ and $z_2^3$ are roots of $z^3$ then we can write the equation as,
$$(z^3-z_1^3)(z^3-z_2^3)=0$$
and by expanding the equation we have,
$$z^6-(z_1^3+z_2^3)z^3+z_1^3z_2^3=0$$
So we finally have,
$$z^6+qz^3-\frac{p^3}{27}=0$$
$z_1=\sqrt[3]{-\frac q2+\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}}$
$z_2=\sqrt[3]{-\frac q2-\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}}$
$$\begin{bmatrix} y_1 \\ y_2 \\ y_3 \\ \end{bmatrix}=\begin{bmatrix} z_1 \\ z_2 \\ \end{bmatrix} * \begin{bmatrix} 1 & 1 \\ \omega & \omega^2 \\ \omega^2 & \omega \\ \end{bmatrix}$$
$\omega$ is imaginary cubic root of unity.
$x_j=y_j-\frac{a_2}{3a_3} , j=1,2,3$

For the quartic equation, I will rewrite it as,
$$y^4+py^2+qy+r=0 \tag 3$$
Let $y=z_1+z_2+z_3$
By squaring both sides
$y^2=z_1^2+z_2^2+z_3^2+2(z_1 z_2+z_1 z_3+z_2 z_3)$
$y^2-(z_1^2+z_2^2+z_3^2)=2(z_1 z_2+z_1 z_3+z_2 z_3)$
By squaring both sides again.
$y^4-2(z_1^2+z_2^2+z_3^2 ) y^2+(z_1^2+z_2^2+z_3^2 )^2=4(z_1^2 z_2^2+z_1^2 z_3^2+z_2^2 z_3^2+2(z_1^2 z_2 z_3+z_1 z_2^2 z_3+z_1 z_2 z_3^2 ))$

$y^4-2(z_1^2+z_2^2+z_3^2 ) y^2+(z_1^2+z_2^2+z_3^2 )^2=4(z_1^2 z_2^2+z_1^2 z_3^2+z_2^2 z_3^2)+8z_1 z_2 z_3 (z_1+z_2+z_3 )$

$$y^4-2(z_1^2+z_2^2+z_3^2 ) y^2-8(z_1 z_2 z_3)y+(z_1^2+z_2^2+z_3^2 )^2-4(z_1^2 z_2^2+z_1^2 z_3^2+z_2^2 z_3^2 )=0 \tag 4$$

By comparing the coefficients of $eq(4)$ with the original equation (3) we get,
$p=-2(z_1^2+z_2^2+z_3^2 )$
$q=-8(z_1 z_2 z_3 )$
$r=(z_1^2+z_2^2+z_3^2 )^2-4(z_1^2 z_2^2+z_1^2 z_3^2+z_2^2 z_3^2 )$
and by some little steps of substitution we get,
$z_1^2+z_2^2+z_3^2 =-\frac p2$
$z_1^2 z_2^2 z_3^2=\frac {q^2}{64}$
$z_1^2 z_2^2+z_1^2 z_3^2+z_2^2 z_3^2=\frac{p^2-4r}{16}$
assume that $z_1^2$ , $z_2^2$ and $z_3^2$ are roots of $z^2$ then we can write the equation as,
$$(z^2-z_1^2)(z^2-z_2^2)(z^2-z_3^2)=0$$
and by expanding the equation we have,
$$z^6-(z_1^2+z_2^2+z_3^2 ) z^4+(z_1^2 z_2^2+z_1^2 z_3^2+z_2^2 z_3^2 ) z^2-z_1^2 z_2^2 z_3^2=0$$
By substitution we get a cubic equation,
$$z^6+\frac p2 z^4+\left(\frac{p^2-4r}{16}\right) z^2-\frac{q^2}{64}=0$$
So we finally have,
$$\begin{bmatrix} y_1 \\ y_2 \\ y_3 \\ y_4 \\ \end{bmatrix}=\begin{bmatrix} z_1 \\ z_2 \\ z_3 \\ \end{bmatrix} * \begin{bmatrix} 1 & 1 & 1 \\ 1 & -1 & -1 \\ -1 & 1 & -1 \\  -1 & -1& 1 \\ \end{bmatrix}$$
$x_j=y_j-\frac{a_3}{4a_4} , j= 1,2,3,4$

For higher degree equations,

There is no formula for higher degree equations in radicals
Abel–Ruffini theorem  and Galois theory have proved the impossibility of solving quintic  and higher degree equations in radicals.