Let $a, b, c$ be the lengths of three sides of a triangle satistying the condition $\left(a^2+b^2\right) x^2-2 b(a+c) x+\left(b^2+c^2\right)=0$. If the set of all possible values of $x$ is the interval $(\alpha, \beta)$, then $12\left(\alpha^2+\beta^2\right)$ is equal to __________.

<p>$\left(a^2+b^2\right) x^2-2 b(a+c) x+b^2+c^2=0$</p> <p>$$\begin{aligned} & \Rightarrow a^2 x^2-2 a b x+b^2+b^2 x^2-2 b c x+c^2=0 \\ & \Rightarrow(a x-b)^2+(b x-c)^2=0 \\ & \Rightarrow a x-b=0, \quad b x-c=0 \\ & \Rightarrow a+b>c \quad b+c>a \quad c+a>b \end{aligned}$$</p> <p>$$\begin{array}{l|l|l} a+a x>b x & a x+b x>a & a x^2+a>a x \\ a+a x>a x^2 & a x+a x^2>a & x^2-x+1>0 \\ x^2-x-1<0 & x^2+x-1>0 & \text { always true } \end{array}$$</p> <p>$$\begin{aligned} & \frac{1-\sqrt{5}}{2}< x<\frac{1+\sqrt{5}}{2} \\ & x< \frac{-1-\sqrt{5}}{2}, \text { or } x >\frac{-1+\sqrt{5}}{2} \end{aligned}$$</p> <p>$$\begin{aligned} & \Rightarrow \frac{\sqrt{5}-1}{2}< x<\frac{\sqrt{5}+1}{2} \\ & \Rightarrow \alpha=\frac{\sqrt{5}-1}{2}, \beta=\frac{\sqrt{5}+1}{2} \\ & 12\left(\alpha^2+\beta^2\right)=12\left(\frac{(\sqrt{5}-1)^2+(\sqrt{5}+1)^2}{4}\right)=36 \end{aligned}$$</p>