What does it Equal? Part 2

$x^2+x^{-2}=3$

$x^2+\frac{1}{x^2}=3 \rightarrow x^2=q$

$q+\frac{1}{q}=3$

$q-3+\frac{1}{q}=0$

$q^2-3q+1=0$

$q=\frac{3 \pm \sqrt{5}}{2}$

$x=\pm \sqrt{\frac{3 \pm \sqrt{5}}{2}}$

$a=3$

$b=1$

$c=5$

$d=2$

$3+1+5+2+2=\boxed{13}$

$\begin{aligned} x^2 + x^{-2} & = 3 \\ x^2 + \frac 1{x^2} & = 3 & \small \color{#3D99F6} \text{Add } -2 \text{ on both sides.} \\ x^2 - 2 + \frac 1{x^2} & = 1 \\ \left(x - \frac 1x\right)^2 & = 1 & \small \color{#3D99F6} \text{Take square root on both sides.} \\ x - \frac 1x & = \pm 1 \end{aligned}$

$\implies \begin{cases} x^2 - x - 1 & = 0 & \implies \begin{cases} x = \dfrac {1+\sqrt 5}2 = \sqrt{\left(\dfrac {1+\sqrt 5}2 \right)^2} = \sqrt{\dfrac {3+\sqrt 5}2} \\ x = \dfrac {1-\sqrt 5}2 = - \sqrt{\left(\dfrac {\sqrt 5-1}2 \right)^2} = - \sqrt{\dfrac {3-\sqrt 5}2} \end{cases} \\ x^2 + x - 1 & = 0 & \implies \begin{cases} x = \dfrac {-1+\sqrt 5}2 = \sqrt{\left(\dfrac {\sqrt 5-1}2 \right)^2} = \sqrt{\dfrac {3-\sqrt 5}2} \\ x = \dfrac {-1-\sqrt 5}2 = - \sqrt{\left(\dfrac {1+\sqrt 5}2 \right)^2} = - \sqrt{\dfrac {3+\sqrt 5}2} \end{cases} \end{cases}$

Therefore, $a+b+c+d+2 = 3+1+5+2+2 = \boxed{13}$ .