Question 102279
In the standard form 
{{{ax^2 +bx +c = 0  }}}
the roots are:{{{x=-b +- sqrt (b^2 -4*a*c )) / (2*a)}}}
if 	
{{{(b^2-4ac)>= 0}}}
 then the roots are {{{real}}}.  Otherwise, the roots are {{{imaginary}}}  

So
(a)	{{{(b^2 - 4*3*3)>= 0}}}
        {{{ b^2 >=36}}}
        {{{ b>= 6 }}}  the roots are {{{real}}} for any vale of {{{b >= 6}}}

(b)	{{{(b^2 - 4*5*1)>= 0}}}
        {{{b^2 >=(4*5)}}}
        {{{b >= sqrt(4*5)}}}
        {{{b >= 2* sqrt(5)}}}  the roots are {{{real}}}for any vale of {{{b >= 2* sqrt (5)}}}

(c)	{{{(b^2 - (4*-3*-3))>= 0}}}
        {{{b^2 >= 36}}}
        {{{b >= sqrt (36)}}}
        {{{b >=  6}}}    the roots are {{{real}}} for any value of {{{b >= 6}}}