The set of all real numbers x for which x2-x+2+x>0, is

Case I: When x+2≥0 i.e. x≥-2,Then given inequality becomesx2-(x+2)+x>0⇒x2-2>0⇒x>2⇒x<-2 or x>2As x≥-2, therefore, in this case the part of the solution set is [-2, -2]∪(2, ∞).Case II: When x+2≤0 i.e., x≤-2,Then given inequality becomes x2+(x+2)+x>0⇒x2+2x+2>0⇒(x+1)2+1>0, which is true for all real xHence, the part of the solution set in this case is (-∞, -2]. Combining the two cases, the solution set is (-∞,-2)∪([-2,-2]∪(2,∞)=(-∞, -2)∪(2,∞).