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

Aug 18, 2016

$r = \sqrt{2}$ and $r = - \sqrt{3}$

#### Explanation:

${r}^{2} + \left(\sqrt{3} - \sqrt{2}\right) r = \sqrt{6} \to {r}^{2} + \left(\sqrt{3} - \sqrt{2}\right) r - \sqrt{2} \sqrt{3} = 0$

then

$\left(r - \sqrt{2}\right) \left(r + \sqrt{3}\right) = 0$ and the solutions are

$r = \sqrt{2}$ and $r = - \sqrt{3}$