Thrice the square of a natural number decreased by 4 times the number is equal to 50 more than the number. The number is  ?

Let's denote the natural number as "n".

According to the problem, we can write the equation as follows:

3n^2 - 4n = n + 50

Simplify the equation:

3n^2 - 4n - n - 50 = 0

Combine like terms:

3n^2 - 5n - 50 = 0

This is a quadratic equation in the form of ax^2 + bx + c = 0, where a = 3, b = -5, and c = -50.

We can solve this equation using the quadratic formula, which is x = [-b ± sqrt(b^2 - 4ac)] / 2a.

Substitute a, b, and c into the formula:

n = [5 ± sqrt((-5)^2 - 43(-50))] / 2*3 n = [5 ± sqrt(25 + 600)] / 6 n = [5 ± sqrt(625)] / 6 n = [5 ± 25] / 6

So the solutions are n = 30/6 = 5 and n = -20/6 = -10/3.

However, since n is a natural number, it cannot be negative or a fraction. Therefore, the natural number we are looking for is 5.