Question

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

Answer 1

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

Answer 2

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