None of the given point even lie on the parabola (curved line) of the graph, so they can't possibly be zeros. The only explanation is that there simply aren't any zeros.
[tex]x^2+4x+5=0\qquad\text{subtract 5 from both sides}\\\\x^2+2(x)(2)=-5\qquad\text{add}\ 2^2\ \text{to both sides}\\\\x^2+2(x)(2)+2^2=-5+2^2\qquad\text{use}\ (a+b)^2=a^2+2ab+b^2\\\\(x+2)^2=-5+4\\\\(x+2)^2=-1<0\\\\\text{Conclusion:}\\\text{No solutions}\\\\\text{Therefore your answer is B.}[/tex]
