Question 1088372
{{{mx-7=-m(x+1)^2-5}}}

{{{-m(x^2+2x+1)-5-mx+7=0}}}

{{{-mx^2-2mx-m-mx+2=0}}}

{{{-mx^2-3mx-m+2=0}}}

{{{mx^2+3mx+m-2=0}}}


ONE intersection point: {{{(3m)^2-4*m*(m-2)=0}}}
{{{9m^2-4m^2+8m=0}}}
{{{5m^2+8m=0}}}
{{{m(5m+8)=0}}}, 
and because given {{{m<0}}}, need to have {{{5m+8=0}}}
{{{5m=-8}}}
{{{highlight(m=-8/5)}}}.