Question 166360
Consecutive integers follow the pattern: x, x+1, x+2, x+3...


Since the "sum of the squares of six consecutive integers is 1111", this means that {{{x^2+(x+1)^2+(x+2)^2+(x+3)^2+(x+4)^2+(x+5)^2=1111}}}



Now FOIL the expressions to get


{{{x^2+(x^2+2x+1)+(x^2+4x+4)+(x^2+6x+9)+(x^2+8x+16)+(x^2+10x+25)=1111}}}



and combine like terms to get



{{{6x^2+30x+55=1111}}}



I'll let you solve the equation.