1. **State the problem:** Solve the radical equation $$20 - 3\sqrt{t} = \sqrt{t} - 4$$. 2. **Isolate the radical terms:** Move all terms involving $\sqrt{t}$ to one side and constants to the other: $$20 + 4 = \sqrt{t} + 3\sqrt{t}$$ $$24 = 4\sqrt{t}$$ 3. **Simplify the equation:** Divide both sides by 4: $$\frac{24}{\cancel{4}} = \frac{4\sqrt{t}}{\cancel{4}}$$ $$6 = \sqrt{t}$$ 4. **Square both sides to solve for $t$:** $$6^2 = (\sqrt{t})^2$$ $$36 = t$$ 5. **Check for extraneous solutions:** Substitute $t=36$ back into the original equation: Left side: $$20 - 3\sqrt{36} = 20 - 3 \times 6 = 20 - 18 = 2$$ Right side: $$\sqrt{36} - 4 = 6 - 4 = 2$$ Both sides equal, so $t=36$ is a valid solution. **Final answer:** $$t = 36$$