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.

Use the following substitution.  $$x^2+18x+45=t$$ and we obtain
$$t-2\sqrt{t}-15=0$$ ot
$$(\sqrt{t}-1)^2=16$$ and since $\sqrt{t}\geq0$, we obtain $$\sqrt{t}=5,$$
$$x^2+18x+20=0$$ or
$$(x+9)^2=61$$ and we get the answer:
$$\{-9\pm\sqrt{61}\}$$
