Question

(3х-5)(х+2)=(х+4)^2-28

Answer 1

$(3x -5)(x + 2) = (x + 4)^2 - 28\\\\ 3x^2 + 6x -5x - 10 = x^2 + 8x + 16 - 28\\\\ 3x^2 + x - 10 = x^2 + 8x - 12\\\\ 2x^2 - 7x + 2 = 0 \ | \ : 2 \\\\ x^2 -\frac{7}{2}x + 1 = 0$

По теореме Виета:

$x_1 + x_2 = -b\\\\ x_1 + x_2 = \frac{7}{2} = \boxed{3.5}$

Через дискриминант можно найти так:

$2x^2 - 7x + 2 = 0\\\\ D = (-7)^2 - 4*2*2 = 49 - 16 = 33\\\\ x_1 = \frac{7 - \sqrt{33}}{4}\\\\ x_2 = \frac{7 + \sqrt{33}}{4}\\\\ x_1 + x_2 = \frac{7 - \sqrt{33}}{4} + \frac{7 + \sqrt{33}}{4} = \frac{7}{4} - \frac{\sqrt{33}}{4} + \frac{7}{4} + \frac{\sqrt{33}}{4}= \frac{2*7}{4} = \frac{7}{2} = \boxed{3.5}$