Find three consecutive natural numbers such that the product of the second and third

Find three consecutive natural numbers such that the product of the second and third of these numbers is 50 more than the square of the first.

Let us denote the first number from a given sequence of three consecutive natural numbers by x.

Then the second and third numbers from this sequence will be equal to x + 1 and x + 2, respectively.

According to the condition of the problem, the product of the second and third of these numbers is 50 more than the square of the first number, therefore, we can compose the following equation:

(x + 1) * (x + 2) = x² + 50.

We solve the resulting equation:

x² + 2x + x + 2 = x² + 50;

x² + 3x + 2 = x² + 50;

x² + 3x – x² = 50 – 2;

3x = 48;

x = 48/3 *

x = 16.

Knowing the first number, we find the second and third:

x + 1 = 17;

x + 2 = 18.

Answer: the required numbers are 16, 17 and 18.