Root3minus 5root2(root3+root2)=find when a+b=root6 find a and b

$$-5\sqrt{2}(\sqrt{3} + \sqrt{2}) = (-5\sqrt{2} \times \sqrt{3}) + (-5\sqrt{2} \times \sqrt{2})$$ $$= -5\sqrt{2 \times 3} - 5 \times (\sqrt{2})^2$$ $$= -5\sqrt{6} - 5 \times 2$$ $$= -5\sqrt{6} - 10$$

$$\sqrt{3} - 5\sqrt{6} - 10$$

This expression cannot be simplified further because $\sqrt{3}$, $\sqrt{6}$, and the constant term $-10$ are not like terms.

The simplified expression is $\boxed{\sqrt{3} - 5\sqrt{6} - 10}$.

Regarding the second part of your question, "find when $a+b=\sqrt{6}$ find $a$ and $b$":
The equation $a+b=\sqrt{6}$ has infinitely many possible solutions for $a$ and $b$. For example, if $a=1$, then $b=\sqrt{6}-1$. If $a=0$, then $b=\sqrt{6}$. Without additional information or another equation relating $a$ and $b$, it is not possible to find unique values for $a$ and $b$.