Let the two numbers be \( x \) and \( y \). According to the problem, we have the following two equations:
- \( x + y = 22 \)
- \( x \cdot y = 120 \)
From the first equation, we can express \( y \) in terms of \( x \):
\[ y = 22 - x \]
Now, we can substitute \( y \) in the second equation:
\[ x(22 - x) = 120 \]
Expanding this, we get:
\[ 22x - x^2 = 120 \]
Rearranging the equation results in:
\[ x^2 - 22x + 120 = 0 \]
Next, we can factor this quadratic equation. We need two numbers that multiply to 120 and add up to -22. These numbers are -12 and -10:
\[ (x - 12)(x - 10) = 0 \]
Setting each factor to zero gives us the solutions:
\[ x - 12 = 0 \quad \Rightarrow \quad x = 12 \] \[ x - 10 = 0 \quad \Rightarrow \quad x = 10 \]
Thus, the two numbers are 10 and 12. Since we need to input the lesser number first, the answer is:
10, 12
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