Precalculus (10th Edition)

$x^2+2x+4>0$, I complete the square: $(x+1)^2+3>0$, $(x+1)^2\geq0$, because all squares of real numbers are non-negative, hence $(x+1)^2+3\geq0+3=3\gt0$ hence the solution set is the real numbers. (also see the graph of ( $(x+1)^2+3$)