Question

Решите уравнение (3х-4)(х+2)=(х-3)^2-19

Answer 1

Ответ:

$3 {x}^{2} + 6x - 4x - 8 = {x}^{2} - 6x \\ + 9 - 19 \\ 3 {x}^{2} + 2x - 8 = {x}^{2} - 6x - 10 \\ 3 {x}^{2} + 2x - 8 - x {}^{2} + 6x + 10 = 0$

$2 {x}^{2} + 8x + 2 = 0 \div 2 \\ {x}^{2} + 4x + 1 = 0$

$d = 16 - 4 = 12 = 2 \sqrt{3}$

$x1 = \frac{ - 4 - 2 \sqrt{3} }{2} = - 2 - 2 \sqrt{3}$

$x2 = \frac{ - 4 + 2 \sqrt{3} }{2} = - 2 + 2 \sqrt{3}$