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