г) 4m² - 2m - 3 = 2;

Преобразуем уравнение: $$4m^2 - 2m - 3 - 2 = 0$$ $$4m^2 - 2m - 5 = 0$$ Решаем квадратное уравнение вида $$ax^2 + bx + c = 0$$: $$D = b^2 - 4ac = (-2)^2 - 4 \cdot 4 \cdot (-5) = 4 + 80 = 84$$ $$m_1 = \frac{-b + \sqrt{D}}{2a} = \frac{2 + \sqrt{84}}{2 \cdot 4} = \frac{2 + \sqrt{84}}{8} = \frac{2 + 2\sqrt{21}}{8} = \frac{1 + \sqrt{21}}{4}$$ $$m_2 = \frac{-b - \sqrt{D}}{2a} = \frac{2 - \sqrt{84}}{2 \cdot 4} = \frac{2 - \sqrt{84}}{8} = \frac{2 - 2\sqrt{21}}{8} = \frac{1 - \sqrt{21}}{4}$$ Ответ: $$m_1 = \frac{1 + \sqrt{21}}{4}$$, $$m_2 = \frac{1 - \sqrt{21}}{4}$$