Solving $x^2 + 18x + 30 = 2 \sqrt{x^2 + 18x + 45}$? (1983 AIME problem 3) 
What is the product of the real roots of the equation $x^2 + 18x + 30 = 2 \sqrt{x^2 + 18x + 45}$?

I know it is a messy/bad idea, but I first started off by squaring both sides and moving everything to one side to get $$x^4 + 36x^3 + 384x^2 + 1080x + 900 - 4x^2 - 72x - 180 = x^4 + 36x^3 + 380x^2 + 1008x + 720 .$$ 
And by (generalisation) of Vieta's formula, the product of the real roots should be $\frac{720}{1} = 720$, but that is wrong, and I don't understand why.

$$x^2 + 18x + 30 = 2 \sqrt{x^2 + 18x + 45}$$
let $x^2+18x+30=u$
so
$$u= 2 \sqrt{u+15}$$
$$u^2=4(u+15)$$
$$u^2-4u-60=0$$
$$(u+6)(u-10)=0$$
so
$$({\color{Red} {x^2+18x+36}})({\color{Blue} {x^2+18x+20}})=0$$
the roots of $({\color{Red} {x^2+18x+36}})$ do not satisfy the original equation especially in the term $\sqrt{x^2 + 18x + 45}$ so the product of real roots will be $20$ 
