Inspired by Daniel Liu

Let's start with the Vieta's formula

We then know that $\displaystyle \alpha + \beta + \gamma = \frac{3}{2}$

$\displaystyle \sin(\frac{\cos^{-1}(\alpha)}{2})=\sqrt{\frac{1-\cos(\cos^{-1}(\alpha))}{2}}$ and is the same for the rest.

then the equation becomes

$\displaystyle \frac{3}{2} = 1+\frac{4}{2\sqrt{2}} \times \sqrt{(1-\alpha)(1-\beta)(1-\gamma)}$

$\displaystyle \frac{1}{8}= (1-\alpha)(1-\beta)(1-\gamma)=1-(\alpha + \beta + \gamma)+(\alpha \beta +\alpha \gamma + \beta \gamma) + \alpha \beta \gamma$ $\displaystyle =1-\frac{3}{2}+\frac{b+d}{8}$

$\displaystyle \frac{5}{8}=\frac{b+d}{8}$

so $\displaystyle b+d= \boxed{5}$

Here we can substitute $\cos { a } =\alpha ,\cos { b } =\beta ,\cos { c } =\gamma \quad$ as $-1\le \alpha ,\beta ,\gamma \le 1$ .

$\therefore \alpha +\beta +\gamma =1+4\sin { \frac { a }{ 2 } } \times \sin { \frac { b }{ 2 } } \times \sin { \frac { c }{ 2 } } =\cos { a } +\cos { b } +\cos { c } \quad$ --Result $1$

Now one observation.Observe that Result $1$ is conditional trigonometrical identity which holds when $a+b+c=180°$ .

By Veita we have $\alpha +\beta +\gamma =\frac { 3 }{ 2 }=\cos { a } +\cos { b } +\cos { c }$ .

$\therefore a=b=c=60°$ (Why?Proof left for reader!)

$\therefore \cos { a } =\cos { b } =\cos { c } =\frac { 1 }{ 2 } =\alpha =\beta =\gamma$ .

Again by Veita we have $\alpha \beta +\gamma \alpha +\beta \gamma =\frac { b }{ 8 } =\frac { 3 }{ 4 }$ and $\alpha \beta \gamma =\frac { -d }{ 8 } =\frac { 1 }{ 8 }$ .

Giving $b=6$ and $d=-1$ .

$\boxed{b+d=5}$ .