$a \alpha^2+b \beta^2+c \alpha \beta+d=0$ is the transformed equation of $4 x^2+\sqrt{3} x y+5 y^2-4=0$
where,
$\begin{aligned}
& \alpha=\frac{\sqrt{3}}{2} x+\frac{y}{2} \\
& \beta=\frac{-x}{2}+\frac{\sqrt{3} y}{2}
\end{aligned}$
From Eqs. (i) and (ii), we get
$x=\frac{\sqrt{3}}{2} \alpha+\frac{\beta}{2} \Rightarrow y=\frac{\alpha}{2}-\frac{\sqrt{3}}{2} \beta$
Putting the value of $x$ and $y$ in $4 x^2+\sqrt{3} x y+5 y^2-4=0$, we get
$\begin{array}{ll}
4\left(\frac{\sqrt{3}}{2} \alpha+\frac{\beta}{2}\right)^2+\sqrt{3}\left(\frac{\sqrt{3}}{2} \alpha+\frac{\beta}{2}\right)\left(\frac{\alpha}{2}-\frac{\sqrt{3}}{2} \beta\right) & +5\left(\frac{\alpha}{2}-\frac{\sqrt{3}}{2} \beta\right)^2-4=0 \\
\Rightarrow 5 \alpha^2+4 \beta^2+\sqrt{3} \alpha \beta-4=0 \\
\therefore a=5, b=4, c=\sqrt{3}, d=-4 \\
\therefore c(a+b+d)=\sqrt{3}(5+4-4)=5 \sqrt{3}
\end{array}$