Let the two consecutive natural numbers be \( n \) and \( n + 1 \). The equation representing their product can be written as:
\[ n(n + 1) = 3782 \]
Expanding this, we have:
\[ n^2 + n - 3782 = 0 \]
This is a quadratic equation in the standard form \( ax^2 + bx + c = 0 \). Here, \( a = 1 \), \( b = 1 \), and \( c = -3782 \).
To find the values of \( n \), we can use the quadratic formula:
\[ n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Plugging in the values:
\[ b^2 - 4ac = 1^2 - 4 \cdot 1 \cdot (-3782) = 1 + 15128 = 15129 \]
Now, calculating the square root:
\[ \sqrt{15129} = 123 \]
Now, substituting back into the quadratic formula:
\[ n = \frac{-1 \pm 123}{2} \]
Calculating the two possible values:
\[ n = \frac{122}{2} = 61 \]
\[ n = \frac{-124}{2} = -62 \quad (\text{not a natural number}) \]
The only valid solution is \( n = 61 \).
Thus, the two consecutive natural numbers are:
\[ 61 \quad \text{and} \quad 62 \]
To check:
\[ 61 \times 62 = 3782 \]
Therefore, the solution to the problem is:
The two consecutive natural numbers are 61 and 62.