Question

Find the smallest of 3 consecutive positive integers such that when 5 times the largest is subtracted from the square of the middle one the result exceeds three times the smallest by 7.

Answer 1

This is how i did it
x, x+1, X+2
(x+1) ^2 - 5 (x+2)=3x+7

Answer 2

x^2+2x+1-5x-10 = 3x+7
x^2-3x-9=3x+7
x^2-6x-16=0
x^2-8x+2x-16 =0
x(x-8)+2(x-8)=0
(x+2)(x-8)=0
x=-2, or 8