x^(2)+x<=6; D=[Reals]

Answer 1

same principle as previous problem

x^2 + x -6 =0

(x+3)(x-2)=0

x= -3. x=2

now test the 3 regions and see which satisfy the original equation

x<-3. (use -4 to test)

x>2 (use 3 to test)

x is between -3 and +2 ( use 0 to test)