The sum of all the real solutions of $(x^2+x{)}^2-8(x^2+x)+12=0$ is:

Let $y=x^2+x$. Then $y^2-8y+12=0$ gives $y=2$ or $y=6$. From $x^2+x-2=0$ the roots are 1 and −2; from $x^2+x-6=0$ the roots are 2 and −3. All four are real, and $1+(-2)+2+(-3)=-2$.