The sum of the square of 3 consecutive positive integers is 110 . what are the numbers

Let your 1st integer be X, then the 2nd will be (X+1) and the 3rd will be (X+2). Since we need the sum of the squares to equal 110 ⇒ X² + (X+1)² + (X+2)² = 110 Expanding the left-hand side: X² + X² + 2·X + 1 + X² +4·X + 4 = 110 3·X² + 6·X + 5 = 110 3·X² + 6·X - 105 = 0 Solve using the quadratic formula and you get roots: X = 5, X = -7 Your problem doesn't state that we have to restrict the solution to positive integers only and since we are summing the squares we have 2 solutions that work: 5, 6, 7 and -7, -6, -5

Aand b is what i think