Question 45269
{{{x^3+6x^2+5x+30=0}}}


This can be solved by factoring by grouping the first two terms and the last two terms together:
{{{x^2(x+6) + 5(x+6) = 0}}}

{{{(x+6)(x^2+5) = 0}}}
{{{x= -6 }}} or {{{x^2= -5}}}


If you are looking for real roots, there is only one, x= -6.


The other solutions would be {{{x^2 = -5}}}, which are NOT real solutions.  If you are solving complex solutions, then {{{x= 0+-sqrt(-5) = 0+-i* sqrt(5) }}} and of course, x= -6.


R^2 at SCC