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.

Write $t=x^2+18x +30$, then we get $$t^2=4t+60$$ so $(t-10)(t+6)=0$ and thus


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*case $x^2+18x+36=0$ so $x_1=x_2=-6$ which is impossible (left side is negative)

*case $x^2+18x+20=0$ so the product of real roots is $20.$
