Question 1096211
{{{n^2+(n+1)^2+(n+2)^2=149}}}

{{{n^2+n^2+2n+1+n^2+4n+4=149}}}

{{{3n^2+6n+5=149}}}

{{{3n^2+6n-144=0}}}

{{{n^2+2n-48=0}}}

{{{(n-6)(n+8)=0}}}


{{{highlight(n=6)}}}, the FIRST of these consecutive natural numbers