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 =0 5x^2 + 45x + 100 = 0 / : 5 x^2 + 9x + 20 = 0 D = 9^2 - 4*1*20 = 81 - 80 = 1 VD = V1 = 1 x1 = -9 +1 / 2 = -8/2 = -4 x2 = -9 - 1 / 2 = - 10/2 = -5
2х^2-3х-104=-3х^2-48х+4 2х^2+3х^2-3х+48х=4+104 5х^2+45х=108 х^2+9х-20=0 а=1; b=9; c=-20 D=b^2-4ac D=1 x1= -b+1=-4 x2=-b-1=-6 Ответ: -4; -6