To complete the expression
\[ \frac{5t^2}{4r} = \frac{\text{nothing}}{12r^4}, \]
we need to find the value for "nothing" such that the two fractions are equivalent.
To do this, we can cross-multiply:
\[ 5t^2 \cdot 12r^4 = 4r \cdot \text{nothing}. \]
This simplifies to:
\[ 60t^2r^4 = 4r \cdot \text{nothing}. \]
Now, we can solve for "nothing" by isolating it:
\[ \text{nothing} = \frac{60t^2r^4}{4r}. \]
Since \(r\) is present in the numerator and the denominator, we can simplify:
\[ \text{nothing} = \frac{60t^2r^3}{4} = 15t^2r^3. \]
So, the complete expression is:
\[ \frac{5t^2}{4r} = \frac{15t^2r^3}{12r^4}. \]
Thus, the final answer for "nothing" is:
\[ 15t^2r^3. \]