Find three consecutive even integers such that the square of the first is 80 less than the square of the third. [Only an algebraic solution will be accepted.]​

First number = 19

Second consecutive = 20

Third consecutive = 21

Step-by-step explanation:

Assume;

First number = x

So,

Second consecutive = x + 1

Third consecutive = x + 1 + 1 = x + 2

So,

(x + 2)² - x² = 80

x² + 2(2)(x) + 2² - x² = 80

x² + 4x + 4 - x² = 80

4x + 4 = 80

4x = 80 - 4

4x = 76

x = 76 / 4

x = 19

First number = x

First number = 19

Second consecutive = x + 1

Second consecutive = 19 + 1

Second consecutive = 20

Third consecutive = x + 2

Third consecutive = 19 + 2

Third consecutive = 21

they can both be solutions depends on the equation

step-by-step explanation:

what is the equation? ?