I have somehow find a way to find solutions of cubic equation by first converting it to quadratic equations .It's a bit embarrassing that I used cubic formula to do so.My question is that is my method correct and if yes can I calculate roots of cubic equation which has complex solutions by this method.So what I have done is Let there be a cubic equation
A$x^3$+B$x^2$+Cx+D=0........(1)
Let there be a quadratic equation which has same solution as equation 1and it's $x^2$coefficient be equal to A i.e the coefficient of $x^3$ in equation 1.So our quadratic equation is
A$x^2$+bx+c=0..........(2)
Writing both equations together and multiplying equation (2) by x we get
A$x^3$+B$x^2$+Cx+D=0........(3)
A$x^3$+b$x^2$+cx=0...........(4)
Subtracting 4 from 3 we get
B$x^2$-b$x^2$+Cx-cx+D=0
$x^2$(B-b)+x(C-c)+D=0............(5)
Now equation 2 and 5 have same solutions and both are quadratic so we can multiply coefficient of 5 by a factor k such that it comes equal to coefficients of equation (2) so
(B-b)k=A...........(6)
(C-c)k=b..........(7)
Dk=c
k=$\frac{c}{D}$..........(8)
Putting value from 8 in 6 and 7 we get
$\frac{(B-b)c}{D}$=A
$\frac{(C-c)c}{D}$=b
c=$\frac{AD}{B-b}$...........(9)
b=$\frac{c(C-c)}{D}$..........(10)
Now what I have done is that I have substituted value of c from (9) in equation (10)and come up with a cubic equation
$b^3$-$b^2$2B+b($B^2$+CA)+$A^2$D-ABC=0.
Now here is the funny and embarrassing part I have calculated value of b from cubic formula to be
b=($(R+$($Q^3$+$R^2$)^$\frac{1}{2}$)^$\frac{1}{3}$)+($(R-$($Q^3$+$R^2$)^$\frac{1}{2}$)^$\frac{1}{3}$)+$\frac{2B}{3}$..............11
where R=$\frac{9ABC-2$B^3$-27$A^2$D}{54}$
Q=$\frac{3CA-$B^2$}{9}$
So now cubic equation will have solutions equal to the solutions of quadratic equation
A$x^2$+bx+c=0 i.e
A$x^2$+bx +$\frac{AD}{B-b}$
and b can be calculated from equation 11,when 2 roots are calculated 3rd root can also be calculated by dividing polynomial .I know this method is quite long but I want to ask is this method correct or not and if correct can this method give solutions of all cubic equations