Play with roots

We can write

$(x-\alpha)(x-\beta)(x-\gamma)(x-\delta)=x^{4}+4x^{3}-6x^{2}+7x-9$

putting $i$ and $-i$ in $x$ and multiplying the equations,

$(i-\alpha)(-i-\alpha)(i-\beta)(-i-\beta)(i-\gamma)(-i-\gamma)(i-\delta)(-i-\delta)=(-2+3i)(-2-3i)$

$(1+\alpha^{2})(1+\beta^{2})(1+\gamma^{2})(1+\delta^{2})=13$

Equation can be written as:- $x^4-6x^2-9=-x(4x^2+7)$ $\implies (x^4-6x^2-9)^2=x^2(4x^2+7)^2\cdots (1)$ $1+\alpha^2=x\implies \alpha=\sqrt{x-1}$ Since $\alpha$ is a root of the equation $(1)$ : $\therefore ((x-1)^2-6(x-1)-9)=(x-1)(4(x-1)+7)^2$ $\implies ((x-1)^2-6(x-1)-9)-(x-1)(4x+3)^2=0$ We have to find the product of roots of this equation since its roots are $1+\alpha^{2},1+\beta^{2},1+\gamma^{2},1+\delta^{2}$ which can be calculated by Vieta's formula and since leading coefficient is $1$ we only have to find the constant term which is given by:- $1+36+81+12-108-18+9=\large\boxed{13}$