What is the product of the [real](https://artofproblemsolving.com/wiki/index.php/Real) [roots](https://artofproblemsolving.com/wiki/index.php/Root) of the [equation](https://artofproblemsolving.com/wiki/index.php/Equation) $x^2 + 18x + 30 = 2 \sqrt{x^2 + 18x + 45}$?
If we were to expand by squaring, we would get a [quartic](https://artofproblemsolving.com/wiki/index.php/Quartic_Equation) [polynomial](https://artofproblemsolving.com/wiki/index.php/Polynomial), which isn't always the easiest thing to deal with.
Instead, we substitute $y$ for $x^2+18x+30$, so that the equation becomes $y=2\sqrt{y+15}$.
Now we can square; solving for $y$, we get $y=10$ or $y=-6$. The second root is extraneous since $2\sqrt{y+15}$ is always non-negative (and moreover, plugging in $y=-6$, we get $-6=6$, which is obviously false). Hence we have $y=10$ as the only solution for $y$. Substituting $x^2+18x+30$ back in for $y$,

$x^2+18x+30=10 \Longrightarrow x^2+18x+20=0.$ Both of the roots of this equation are real, since its discriminant is $18^2 - 4 \cdot 1 \cdot 20 = 244$, which is positive. Thus by [Vieta's formulas](https://artofproblemsolving.com/wiki/index.php/Vieta%27s_formulas), the product of the real roots is simply $\boxed{020}$.