Question 937988
What happens if you put the function into a graph?


You can try the old slow way of making a data table for getting several points and then plotting them, and try to guess the ROOTS or ZEROS.


{{{graph(300,300,-6,6,-6,6,x^2+2x+3)}}}


This parabola does not cross nor touch the x-axis, and therefore has NO REAL ZEROS.  




But the function DOES have Complex zeros.  
Accept a number, i such that:
{{{highlight_green(i^2=-1)}}}.
Raising this to the {{{1/2}}} power, you find that {{{i=sqrt(-1)}}}  or  {{{i=-sqrt(-1)}}}.



NEXT,  set f(x) to 0, and SOLVE FOR x, using Completing the Square.
{{{x^2+2x+3=0}}}
{{{x^2+2x+(2/2)^2+3=(2/2)^2}}},  the full expression shown so it will be less of a mystery
{{{x^2+2x+1+3=1}}}
{{{(x^2+2x+1)+3=1}}}
{{{(x+1)^2+3=1}}}

{{{(x+1)^2=-2}}}

{{{x+1=0+- sqrt(-2)}}}

{{{x=-1+- sqrt(-2)}}}

{{{x=-1+- sqrt(-1*2)}}}

{{{x=-1+- sqrt(-1)sqrt(2)}}}

{{{highlight(highlight(x=-1+- i*sqrt(2)))}}}