Let the numbers be \( x \) and \( y \). The equations are:
\[\nx + y = 11 \quad \text{and} \quad x \times y = 30\n\]
Solve using a quadratic. Substitute \( y = 11 - x \):
\[\nx \times (11 - x) = 30\n\]
\[\nx^2 - 11x + 30 = 0\n\]
Factor the quadratic:
\[\n(x - 6)(x - 5) = 0\n\]
So, \( x = 6 \) and \( y = 5 \) (or vice versa). The answer is \( 6, 5 \).