Question

ПОЖАЛУЙСТА ПОМОГИТЕ!!

Answer 1

1.

$- {x}^{2} = 2x - 3$

$2x - 3 + {x}^{2} = 0$

a = 1, b = 2, c = –3

$D=b²-4ac = {2}^{2} - 4 \times 1 \times ( - 3) = 4 + 12 = 16 = {4}^{2}$

тогда

$x1 = \frac{ - b + \sqrt{D} }{2a} = \frac{ - 2 + 4}{2} = \frac{2}{2} = 1$

$x2 = \frac{ - b - \sqrt{D} }{2a} = \frac{ - 2 - 4}{2} = \frac{ - 6}{2} = - 3$

2.

$0.5 {x}^{2} = 2x$

$0.5 {x}^{2} - 2x = 0$

$x(0.5x - 2) = 0$

$x1 = 0$

или

$0.5x - 2 = 0$

$0.5x = 2$

$x2= \frac{2}{0.5} = 4$