Proof that given equation(quartic) doesn't have real roots $$
(x^2-9)(x-2)(x+4)+(x^2-36)(x-4)(x+8)+153=0
$$
I need to prove that the above equation doesn't have a real solution. I tried breaking it up into an $(\alpha)(\beta)\cdots=0$ expression, but no luck. Wolfram alpha tells me that the equation doesn't have real roots, but I'm sure there's simpler way to solve this than working trough the quartic this gives.

Expand $x^2-9=(x-3)(x+3)$ and $x^2-36=(x-6)(x+6)$. Let $t=x+\frac 1 2$, $u=x+1$. Then
$$LHS=(t^2-(2.5)^2)(t^2-(3.5)^2)+(u^2-25)(u^2-49)+153\geq -(\frac {3.5^2-2.5^2} 2)^2-(\frac {49-25} 2)^2+153=0$$
with equality iff $t^2=\frac {2.5^2+3.5^2} 2$ and $u^2=\frac {25+49} 2$ which is incompatible. Hence $LHS>0$
