Question

решите уравнение:
1.(-2x^2+3x)-(7x-2x^2)=-2
2.-(3x^2-2x+1)+(3x^2-4x-8)=0

Answer 1

1) (- 2x² + 3x) - (7x - 2x²) = - 2
- 2x² + 3x - 7x + 2x² = - 2
- 4x = - 2
x = 0,5

2) - (3x² - 2x + 1) + (3x² - 4x - 8) = 0
- 3x² + 2x - 1 + 3x² - 4x - 8 = 0
- 2x - 9 = 0
- 2x = 9
x = - 4,5

Answer 2

$( - 2x {}^{2} + 3x) - (7x - 2x {}^{2} ) = - 2 \\ - 2x {}^{2} + 3x - 7x + 2x {}^{2} = - 2 \\ - 4x = - 2 \\ x = \frac{1}{2} = 0.5 \\ \\ - (3x {}^{2} - 2x + 1) + (3x {}^{2} - 4x - 8) = 0 \\ - 3x {}^{2} + 2x - 1 + 3x {}^{2} - 4x - 8 = 0 \\ - 2x - 9 = 0 \\ -2x = 9 \\ x = - \frac{9}{2} = - 4.5$