One root of the quadratic equation y^2 + by + c = 0, where b and c are real constants, is 3 - isqrt3. What are b and c?

Note that $3 - i\sqrt3 = 3 - \sqrt{-3}$.

Now, recall the quadratic formula: $x = {-b \pm \sqrt{b^2-4ac} \over 2a}$. Substituting $a = 1$, we have $3 - \sqrt {-3} = {-b - \sqrt{b^2-4c} \over 2}$ (take the minus root, not the plus)

Multiplying both sides by 2 gives us: $6 - 2 \sqrt {-3} = -b - \sqrt{ b^2 - 4ac}$. Simplify the left-hand side: $6 - \sqrt{-12} = -b - \sqrt {b^2 - 4ac}$.

Note that $6 = -b$, meaning $b = -6$. Plugging this into the discriminant, we have ${36 - 4c} = -12$ (disregard the square root because both have them)

Solving, we find that $c = 12$.