Biquadratics gave me Nightmare!

As $x \in \mathbb R$ the range of the function $F(x) = (x-2)(x-3)(x-5)(x-6)$ doesn't depend on translations on the $x$ variable so (just to have a more symmetric problem to deal with) we can solve the equivalent question: how many roots the following equation has?

$G(x) = F(x +4) = (x-2)(x-1)(x+1)(x+2) = 504$

The function $G(x)$ is a fourth degree polynomial with $1$ as coefficient of $x^4$ and with four zeros, namely $-2,-1, 1$ and $2$ .

$G(x)$ is striktly monotone increasing in $x \geq 2$ and here has range $[0, +\infty)$ .

$G(x)$ is striktly monotone decreasing in $x \leq -2$ and here has range $[0, +\infty)$

So there are exactly $2$ values $x_1 \leq -2, x_2 \geq 2$ such that $G(x) = 504$ .

The only other range for $x$ where $G$ is positive is in $x \in (-1,1)$ but here we can use AM-GM and find that

$G(x) = (2-x)(1-x)(x+1)(x+2) \leq \left( \dfrac{6}{4}\right)^4 = \dfrac{81}{16} < 504$

So $x_1, x_2$ are the only roots of $G(x) = 504$ .

Put x^2-8x=y,the equation reduces to (Y+12)(y+15)=504

Y^2+27y-324=0

Implies y=9 or y=-36

When y=9,x^2-8x-9=0 or x=-1 or 9

When y=-36,x^2-8x+36=0,roots are imaginary. Hence only two real roots