Theory of equations

Let the given root be y.
If the equation has 2 equal roots,

f(y)=f'(y)=0

From the above 2 condtions we get,

ay^3+3by^2+3cy+d=0....(1)
ay^2+2by+c=0....(2)

(1)-(2)\times y
\rightarrow by^2+2cy+d=0.......(3)

Solving equations 2 and 3 by determinant method,

\frac{y^2}{2c^2-2bd}=\frac{-y}{bc-ad}=\frac{1}{2b^2-2ac}

\rightarrow y =\frac{bc-ad}{2(ac-b^2)}