Question 43889
Solve:
{{{x^2+2x+5 = 0}}} This will not factor so use the quadratic formula: 
{{{x = (-b+-sqrt(b^2-4ac))/2a}}} where, from your equation:
 a = 1, b = 2, c = 5
{{{x = (-2+-sqrt(2^2-4(1)(5)))/2(1)}}}
{{{x = (-2+-sqrt(4-20))/2}}}
{{{x = (-2+-sqrt(-16))/2}}}
{{{x = -1+2i}}} or {{{x = -1-2i}}}

As you can see, there are solutions to the equation but they are not real solutions, they are complex solutions.