Question

How do you solve `r^2+(sqrt(3)-sqrt(2))r = sqrt(6)`?

Answer 1

`r = sqrt(2)` and `r = -sqrt(3)`

Explanation:

`r^2+(sqrt(3)-sqrt(2))r = sqrt(6)->r^2+(sqrt(3)-sqrt(2))r-sqrt(2)sqrt(3)=0`

then

`(r-sqrt(2))(r+sqrt(3))=0` and the solutions are

`r = sqrt(2)` and `r = -sqrt(3)`