Real roots

$\begin{aligned} x^2+18x + 30 & = 2\sqrt{x^2+18x+45} & \small \color{#3D99F6} \text{Let }y = x^2 + 18x+30 \\ y & = 2\sqrt{y+15} & \small \color{#3D99F6} \text{Squaring both sides} \\ y^2 & = 4(y+15) \\ y^2 - 4y - 60 & = 0 \\ (y-10)(y+6) & = 0 \\ \implies y & = 10 & \small \color{#3D99F6} \text{Note that when }y = -6: -6 \ne 2\sqrt{-6+15} \\ x^2+18x+30 & = 10 \\ x^2+18x+20 & = 0 \end{aligned}$

Note that the roots of $x^2+18x+\color{#3D99F6}20$ are real and by Vieta's formula , their product is $\boxed{\color{#3D99F6}20}$ .

Relevant wiki: Vieta's Formula Problem Solving - Basic

let $y=\sqrt{x^2+18x+45}$ then $y^2=x^2+18x+45$ , we have

$x^2+18x+30=2y$

adding $15$ to both sides, we get

$x^2+18x+45=2y+15$

so,

$y^2-2y-15=0$

$(y-5)(y+3)=0$

$y=5$

$y=-3$ (Rejected since $y$ must be positive.)

Substituting $y=5$ back, we get

$x^2+18x+30=2(5)$

$x^2+18x-20=0$

By Vieta's Formula , the product of the roots is

$x_1x_2=\dfrac{c}{a}=\dfrac{20}{1}=$ $\large{\color{#D61F06}\boxed{20}}$

$x^2 + 18x + 30 = 2\sqrt{x^2 + 18x + 45} \\ x^2 + 18x + 45 - 16 + 1 - 2\sqrt{x^2 + 18x + 45} = 0 \\ (\sqrt{x^2 + 18x + 45})^2 - 2\sqrt{x^2 + 18x + 45} + 1 = 16 \\ (\sqrt{x^2 + 18x + 45} - 1)^2 = 16 \\ \sqrt{x^2 + 18x + 45} - 1 = 4$

(it can't be -4, because the root on the LHS would be negative)

$\sqrt{x^2 + 18x + 45} = 5 \\ x^2 + 18x + 45 = 25 \\ x^2 + 18x +20 = 0$

From Vieta's formula, product of the roots of the final equation is : $\boxed{x_{1}x_2 = 20}$