$\begin{array}{1 1}(A)\;0&(B)\;2\\(C)\;4&(D)\;6\end{array} $

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$x^2+3x+2=(x+1)(x+2)$

$x^2-9x+20=(x-4)(x-5)$

$(x^2+3x+2)(x^2-9x+20)=(x+1)(x-4)(x-2)(x-5)$

$(x^2+3x+2)(x^2-9x+20)=(x^2-3x-4)(x^2-3x-10)$

Now putting $x^2-3x=y$,above equation can be written as,

$(y+1)(y-4)(y-10)=-30$

Roots of this equation are $5,4,\pm\sqrt{30}$

Use these to obtain roots of x

Hence (D) is the correct answer.

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