решите уравнение а)х(х-4)=(х+3) в 2 степени

х(х-4)=(х+3)x² - 4x = x + 3x² - 4x - x - 3 = 0x² - 5x - 3 = 0a=1, b=-5, c=-3D = b² - 4ac = (-5)² - 4*1*(-3) = 25 + 12 = 37;D > 0x1 = -b+√D/2a = 5+√37/2;x2 = -b-√D/2a = 5-√37/2; Ответ : 5+√37/2; 5-√37/2;Если (х+3)² тох(х-4)=(х+3)²x² - 4x = x² + 6x + 9x² - x² -4x - 6x - 9 = 0-10x - 9 = 0 | : (-1)10x + 9 = 010x = -9x = -9/10 = -0,9 Ответ : -0,9