Find three consecutive integers such that the product of the two largest is 20 more than the square of the smallest number

Givens

the smallest integer is x

The second smallest integer is x + 1

The largest integer is x + 2

Equations

(x + 1)(x + 2) = x² + 20

Solution

x² + x + 2x + 2 = x² + 20  Subtract x² from both sides

3x + 2 = 20                       Subtract 2 from both sides.

3x = 20 - 2

3x =  18                              Divide by 3

x = 18/3

x = 6

The three numbers are 6,7 and 8

Check

7*8  = 6^2 +  20

56 = 36 + 20

56 = 56

The equation's solution does check.

17 1/3

step-by-step explanation: