B) (3y²+y-24) / (9-y²) = -2

B) $$ \frac{3y^2 + y - 24}{9 - y^2} = -2 $$

$$ 3y^2 + y - 24 = -2(9 - y^2) $$

$$ 3y^2 + y - 24 = -18 + 2y^2 $$

$$ 3y^2 - 2y^2 + y - 24 + 18 = 0 $$

$$ y^2 + y - 6 = 0 $$

$$ D = b^2 - 4ac = 1^2 - 4 \cdot 1 \cdot (-6) = 1 + 24 = 25 $$

$$ y_1 = \frac{-b + \sqrt{D}}{2a} = \frac{-1 + \sqrt{25}}{2 \cdot 1} = \frac{-1 + 5}{2} = \frac{4}{2} = 2 $$

$$ y_2 = \frac{-b - \sqrt{D}}{2a} = \frac{-1 - \sqrt{25}}{2 \cdot 1} = \frac{-1 - 5}{2} = \frac{-6}{2} = -3 $$

Ответ: $$y_1 = 2, y_2 = -3$$