$\sqrt{x^2 +12y} + \sqrt{y^2 +12x} =33$
$\sqrt{x^2 +12y} - \sqrt{y^2 +12x} =\dfrac{(x-y)(x+y-12)}{33}=\dfrac{x-y}{3}$
$2\sqrt{x^2+12y}=\dfrac{99+x-y}{3}=\dfrac{2(x+38)}{3}$
$9(x^2+12(23-x))=(x+38)^2$
$8(x-10)(x-13)=0$
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