Question 948573
{{{3x^2=k-2x}}}
{{{3x^2+2x-k=0}}}=> {{{a=3}}}, {{{b=2}}}, and {{{c=-k}}}

The argument of the square root, the expression {{{b^2 – 4ac}}}, is called the "discriminant" because, by using its value, you can discriminate between (tell the differences between) the various solution types.

if {{{b^2 – 4ac<0}}} there is no real roots
if {{{b^2 – 4ac=0}}} there is one real roots
if {{{b^2 – 4ac>0}}} there are two real roots

so, we need
if {{{b^2 – 4ac=0}}} there is one real roots
if {{{b^2 – 4ac>0}}} there are two real roots

and I will write both in one as {{{b^2 – 4ac>=0}}}

now just plug in {{{a=3}}}, {{{b=2}}}, and {{{c=-k}}}

{{{2^2 – 4*3*(-k)>=0}}}

{{{4 – 12(-k)>=0}}}

{{{4 +12k>=0}}}

{{{12k>=-4}}}

{{{k>=-4/12}}}

{{{k>=-1/3}}}

so, if {{{k=-1/3}}} we will have one real root which is {{{x=-1/3}}} (you can check it)
if {{{k>=-1/3}}}, using for example {{{k=0}}} we will have two real roots which are {{{x = 0}}} and {{{x=-2/3}}}


so, to get real root/roots, you can use any value for {{{k}}} from [{{{-1/3}}},{{{infinity}}})