Question 526923
Did you mean {{{x^2+2x+4=0}}}?
If that's what you meant,
My previous solution was all wrong. I apologize.
{{{x^2+2x+1=(x+1)^2}}}
so writing the equation as
{{{x^2+2x=-4}}}
and then adding 1 to both sides you get
{{{x^2+2x+1=-3}}}
{{{(x+1)^2=-3}}}
you realize that the only solutions are imaginary numbers.
There are no real number solutions.
Unless you are studying complex numbers, you would say there are no solutions.
In general, quadratic equations like
{{{x^2+2x+4=0}}}
can always be worked on by using the quadratic formula
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
If {{{b^2-4*a*c<0}}}, there are no real number solutions
Other ways to solve quadratic equations is by "completing the square" and by factoring. Graphing can also help.
Completing the square is what I did at the top; I looked for an ending to 
{{{x^2+2x+ending}}}
that would make it a perfect square.
It is the same as using the quadratic formula, except you do not need to remember the formula to use the "complete the square" trick.
Factoring to find
(x+?)(x+??) is based on the fact that if the ??? are a and b
{{{(x+a)(x+b)=(x+a)x+(x+a)b=x^2+ax+bx+ab=x^2+(a+b)x+ab}}}
You look for pairs of factors of the independent term, and figure out which pair of factors adds up to the coefficient for the term in x.
For example, for
{{{x^2-6x+8}}}
the independent term is 8, which is the product of 1 and 8, or -1 and -8, or 2 and 4, or -2 and -4.
Which of those pairs adds up to the -6 in the term -6x?
If you say -2 and -4, you try them and if you get
{{{(x-2)(x-4)=x^2-6x+8}}}
you've found the factoring of {{{x^2-6x+8}}}