Помогите решить уравнения с нахождением x1, x2:
1) 3x2(второй степени) + 13x - 10 = 0
2) 2x2(второй степени) - 3x = 0
3) 2(x+3)(x-1/2) = 0

3x^2 + 13x - 10 = 0D = 169 + 120 = 289 = 17^2 x1 = ( - 13 + 17 ) : 6 = 2/3 x2 = ( - 13 - 17 ) : 6 = 5 ----------------------------------2x^2 - 3x = 0 x( 2x - 3 ) = 0 x1 = 0 2x - 3 = 0 2x = 3 x2 = 1,5 -----------------------------------2( x + 3 )( x - 1/2 ) = 0 x + 3 = 0 x1 = - 3 x - 1/2 = 0 x2 = 1/2 

1) 3x^2+13x-10 = 0D = √(13^2-4*3*(-10) = √(169+120) = √289 = 17x1 = (-13+17)/3*2 = 4/6 = 2/3x2 = (-13-17)/3*2 =-30/6 = -5Ответ: x1 = 2/3, x2 = -52) 2x^2-3x=0x(2x-3) = 0x1 = 0 , x2 = 3/2Ответ: x1 = 0, x2 = 3/23) 2(x+3)(x-1/2)=0(x+3)(2x-1)=0x1=-3, x2=1/2Ответ: x1=-3, x2=1/2