The square of the sum of two consecutive natural numbers is greater than the sum of the squares of these two numbers by 112. find the two numbers.

We solve the equation, (a + a + 1) ^2 = 112 + a^2 + (a + 1) ^2;

Then, (2a + 1) ^2 = 112 + a^2 + a^2 + 2a + 1;

4a^2 + 4a + 1 = 113 + 2a^2 + 2a;

Finally, 2a^2 + 2a - 112 = 0;

a^2 + a - 56 = 0;

We use Quadratic Formula for this Quadratic Equation;

The solutions are a1 = 7 and a2 = - 8;

But a is a natural number; so, a = 7;

The natural consecutive numbers are 7 and 8.