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In the given problem, two legs of a right triangle differ by 4 cm, and the hypotenuse is 20 cm. You need to find the lengths of these two legs.

In a right triangle, according to the Pythagorean theorem, the relationship between the legs \( a \) and \( b \) and the hypotenuse \( c \) is:

\[ a^2 + b^2 = c^2 \]

Given:

• \( c = 20 \)
• \( a = b + 4 \)

Substitute \( a = b + 4 \) into the Pythagorean theorem:

\[ (b + 4)^2 + b^2 = 20^2 \]

\[ b^2 + 8b + 16 + b^2 = 400 \]

Combine like terms:

\[ 2b^2 + 8b + 16 = 400 \]

Subtract 400 from both sides to set the equation to zero:

\[ 2b^2 + 8b + 16 - 400 = 0 \]

\[ 2b^2 + 8b - 384 = 0 \]

Divide the entire equation by 2 to simplify:

\[ b^2 + 4b - 192 = 0 \]

Now solve for \( b \) using the quadratic formula:

\[ b = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A} \]

For this equation, \( A = 1 \), \( B = 4 \), \( C = -192 \):

\[ b = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 1 \cdot (-192)}}{2 \cdot 1} \]

\[ b = \frac{-4 \pm \sqrt{16 + 768}}{2} \]

\[ b = \frac{-4 \pm \sqrt{784}}{2} \]

\[ b = \frac{-4 \pm 28}{2} \]

The solutions are:

\[ b = \frac{-4 + 28}{2} = \frac{24}{2} = 12 \]

\[ b = \frac{-4 - 28}{2} = \frac{-32}{2} = -16 \]

Since a length can't be negative, \( b = 12 \).

Then, \( a = b + 4 = 12 + 4 = 16 \).

So, the two legs are 12 cm and 16 cm.