Find four consecutive integers such that the sum of the squares of the first and second is 65 more than the product of the third and fourth

Question

 -8, -7, -6, and -5 or 9, 10, 11, and 12

-9, -8, -7, and -6 or 12, 13, 14, and 15

-6, -5, -4, and -3 or 11, 12, 13, and 14

-7, -6, -5, and -4 or 10, 11, 12, and 13

Answer 1

The 4 integers are x,x+1,x+2 and x+3

so x^2 + (x+1)^2= (x+2)(x+3) + 65

x^2 + x^2 +2x +1= x^2 +5x +6 +65

x^2 -3x -70 = 0

(x-10)(x+7) =0 so x =10 or -7 so 4 consecutive integers would be 10,11,12and 13 and -7,-6,-5 and -4

which is choice #4