Question 125918
{{{y=x^2+4x}}}
  {{{y=x(x+4)}}}
  {{{y/(x+4) = x}}}
What you have is a relation between
2 variables. You can't "solve" any more 
than you did when you got {{{y = x(x + 4)}}}
If I say {{{y = 10}}}, then there is a 
corresponding value for {{{x}}}, but normally
{{{y}}} depends on whatever {{{x}}} is.
You can find the roots of {{{y=x^2+4x}}}
by setting {{{y = 0}}}. You can see in
{{{y=x(x+4)}}} than {{{x = 0}}} and {{{x = -4}}}
will make {{{y = 0}}}