1. The problem asks to evaluate the expression $$\sqrt{-9} - 6\sqrt{-2}$$ and write the result in the form $a + bi$ where $a$ and $b$ are real numbers. 2. Recall that for any negative number under a square root, we use the imaginary unit $i$ where $$i = \sqrt{-1}$$. 3. Rewrite each square root with $i$: $$\sqrt{-9} = \sqrt{9 \times -1} = \sqrt{9} \times \sqrt{-1} = 3i$$ $$\sqrt{-2} = \sqrt{2 \times -1} = \sqrt{2} \times i = i\sqrt{2}$$ 4. Substitute these back into the expression: $$3i - 6(i\sqrt{2}) = 3i - 6i\sqrt{2}$$ 5. Factor out $i$: $$i(3 - 6\sqrt{2})$$ 6. The expression is now in the form $a + bi$ where $a = 0$ (no real part) and $b = 3 - 6\sqrt{2}$. Final answer: $$0 + (3 - 6\sqrt{2})i$$ So, the real number $a$ equals $0$ and the real number $b$ equals $3 - 6\sqrt{2}$.