find the three consecutive integers such that three times the square of the first is 7 less than twice times the sum of the second and third

Answer 1

The consecutive integers are 1, 2 and 3

Let the 3 integers be x-1, x and x+1

then 3(x-1)^2 = 2(x+x+1) -7

solve for x

3x^2 -6x + 3 = 4x -5

3x^2 -10x +8 = 0

factor

(3x-4)(x-2) = 0

set each factor = 0

3x-4=0

3x=4

x=4/3

ignore this solution as it's not an integer

x-2=0

x=2

the 3 consecutive integers

1, 2, 3

3 times square of 1 is 3

which is 7 less than

2 times (sum of 2 and 3) = 2(5)=10

3 is 7 less than 10