Решите уравнение . 2x^2-3x-104=-3x^2-48x+4

2x^2-3x+104=-3x^2-48x+4 2x^2-3x+104+3x^2+48x-4 =05x^2 + 45x + 100 = 0 / : 5x^2 + 9x + 20 = 0D = 9^2 - 4*1*20 = 81 - 80 = 1VD = V1 = 1x1 = -9 +1 / 2 = -8/2 = -4x2 = -9 - 1 / 2 = - 10/2 = -5

2х^2-3х-104=-3х^2-48х+42х^2+3х^2-3х+48х=4+1045х^2+45х=108х^2+9х-20=0а=1; b=9; c=-20D=b^2-4acD=1x1= -b+1=-4x2=-b-1=-6Ответ: -4; -6