026.3. б) \(\frac{3x+4}{5} - \frac{x^2+4x-3}{3} = 1\)

Умножим обе части уравнения на 15: \[3(3x+4) - 5(x^2 + 4x - 3) = 15\] \[9x + 12 - 5x^2 - 20x + 15 = 15\] \[-5x^2 - 11x + 12 = 0\] \[5x^2 + 11x - 12 = 0\] Дискриминант: \[D = (11)^2 - 4 \cdot 5 \cdot (-12) = 121 + 240 = 361 = 19^2\] Корни: \[x_1 = \frac{-11 + 19}{10} = \frac{8}{10} = 0.8\] \[x_2 = \frac{-11 - 19}{10} = \frac{-30}{10} = -3\] Ответ: \(x_1 = 0.8\), \(x_2 = -3\)