2. приведите уравнение (х-5)(2х+3)-(х+4)(4-х)=2х+5 к виду ах^2+bх+с=0 A. 3x^2+5x-26=0 b. 6x^2-5x+12=0

(х - 5)(2х + 3) - (х + 4)(4 - х) = 2х + 5,

2х² 10х + 3х 15 (16 х²) = 2х + 5,

2х² 10х + 3х 15 16 + х² 2х 5 = 0,

3х² 9х 20 = 0.

D = b² - 4ac

D = 81 - 4 * 3 * (-20) = 81 + 240 = 321.

х = (-b D) / 2a,

х = (9 321) / 6,

х₁ = (9 - 321) / 6, х₂ = (9 + 321) / 6.

А. 3х² + 5x 26 = 0,

D = b² - 4ac

D = 25 - 4 * 3 * (-26) = 25 + 312 = 337.

х = (-b D) / 2a,

х = (-5 337) / 6,

х₁ = (-5 - 337) / 6, х₂ = (-5 + 337) / 6.

В. 6x² - 5x + 12 = 0,

D = b² - 4ac

D = 25 - 4 * 6 * 12 = 25 - 288 = -263.

Уравнение не имеет корней.

С. x² - 9x 12 = 0,

D = b² - 4ac

D = 81 - 4 * (-12) = 81 + 48 = 129.

х = (-b D) / 2a,

х = (9 129) / 2,

х₁ = (9 - 129) / 2, х₂ = (9 + 129) / 2.

D. x²  - 3x 12 = 0,

D = b² - 4ac

D = 9 - 4 * (-12) = 9 + 48 = 57.

х = (-b D) / 2a,

х = (3 57) / 2,

х₁ = (3 - 57) / 2, х₂ = (3 + 57) / 2.