$Thesumofthenon-realrootsof(p^2+p-3)(p^2+p-2)-12=0$

$\begin{array}{l}GivenEquation(p^2+p-3)(p^2+p-2)-12=0\\ Let{p}^2+p-2=y\\ \Rightarrow y(y-1)-12=0\\ \Rightarrow y^2-y-12=0\\ \Rightarrow y^2-4p+3p-12=0\\ \Rightarrow 9y-4(y+3)=0\\ \Rightarrow y=4,-3\\ Ify=4then{p}^2+p-2=4\\ \Rightarrow p^2+p-6=0\\ \Rightarrow p^2+3p-2p-6=0\\ \Rightarrow (p+3)(p-2)=0\\ p=-3,2\\ Ify=-3then{p}^2+p-2=-3\\ \Rightarrow p^2+p+1=0\\ D=1-4=-3<0\\ Norealroots\\ \therefore Sumofnonrealvalues=-3+2=-1\end{array}$