Let the two numbers be \( x \) and \( y \). We can set up the problem with the following equations based on the information provided:
- \( x + y = 15 \) (the sum of the numbers)
- \( x \cdot y = 56 \) (the product of the numbers)
From the first equation, we can express \( y \) in terms of \( x \):
\[ y = 15 - x \]
Now, we can substitute this expression for \( y \) into the second equation:
\[ x \cdot (15 - x) = 56 \]
Expanding this gives:
\[ 15x - x^2 = 56 \]
Rearranging the equation leads to:
\[ x^2 - 15x + 56 = 0 \]
Next, we can solve the quadratic equation using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 1 \), \( b = -15 \), and \( c = 56 \).
Calculating the discriminant:
\[ b^2 - 4ac = (-15)^2 - 4 \cdot 1 \cdot 56 = 225 - 224 = 1 \]
Now substituting back into the quadratic formula:
\[ x = \frac{15 \pm \sqrt{1}}{2} = \frac{15 \pm 1}{2} \]
This gives us two possible values for \( x \):
- \( x = \frac{16}{2} = 8 \)
- \( x = \frac{14}{2} = 7 \)
Using \( x = 8 \):
\[ y = 15 - 8 = 7 \]
Using \( x = 7 \):
\[ y = 15 - 7 = 8 \]
Thus, the two numbers are \( 7 \) and \( 8 \).
In conclusion, the numbers are 7 and 8.