г) (2y-5) / (y+5) = (3y+21) / (2y-1)

г) $$ \frac{2y-5}{y+5} = \frac{3y+21}{2y-1} $$

$$ (2y-5)(2y-1) = (3y+21)(y+5) $$

$$ 4y^2 -2y -10y + 5 = 3y^2 + 15y + 21y + 105 $$

$$ 4y^2 - 12y + 5 = 3y^2 + 36y + 105 $$

$$ 4y^2 - 3y^2 - 12y - 36y + 5 - 105 = 0 $$

$$ y^2 - 48y - 100 = 0 $$

$$ D = b^2 - 4ac = (-48)^2 - 4 \cdot 1 \cdot (-100) = 2304 + 400 = 2704 $$

$$ y_1 = \frac{-b + \sqrt{D}}{2a} = \frac{48 + \sqrt{2704}}{2 \cdot 1} = \frac{48 + 52}{2} = \frac{100}{2} = 50 $$

$$ y_2 = \frac{-b - \sqrt{D}}{2a} = \frac{48 - \sqrt{2704}}{2 \cdot 1} = \frac{48 - 52}{2} = \frac{-4}{2} = -2 $$

Ответ: $$y_1 = 50, y_2 = -2$$