Question 1092480

 if the discriminant {{{b^2 – 4ac>0}}}, the equation {{{x^2-2kx+k^2-2k=6}}} will have two real roots

if the discriminant {{{b^2 – 4ac=0}}}, the equation {{{x^2-2kx+k^2-2k=6}}} will have one real root

if the discriminant {{{b^2 – 4ac<0}}}, the equation {{{x^2-2kx+k^2-2k=6}}} will have no real roots

so, in your case we have

{{{x^2-2kx+k^2-2k=6}}}
{{{x^2-2kx+(k^2-2k-6)=0}}}->{{{a=1}}},{{{b=2k}}},{{{c=k^2-2k-6}}}

{{{b^2 – 4ac=0}}}

{{{(2k)^2 – 4*1(k^2-2k-6)=0}}}

{{{4k^2 – 4k^2+8k+24=0}}}

{{{8k+24=0}}}

{{{8k=-24}}}

{{{k=-24/8}}}

{{{k=-3}}}-> for this value of {{{k}}} your equation will have one real solution

{{{x^2-2(-3)x+(-3)^2-2(-3)-6=0}}}

{{{x^2+6x+9+6-6=0}}}

{{{x^2+6x+9=0}}}

{{{ graph( 600, 600, -10, 10, -10, 10, x^2+6x+9) }}}


and, first integer greater than {{{-3}}} is {{{-2}}}

if {{{k=-2}}} your equation will have two real solutions

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

{{{x^2+4x+4+4-6=0}}}

{{{x^2+4x+2=0}}}

{{{ graph( 600, 600, -10, 10, -10, 10, x^2+4x+2) }}}