Question

найдите корни уравнения x^2+4x+3=0​

Answer 1

D =4^2-4*1*3=16-12=4

x1=-4+√4/2*1=-1

x2=-4-√4/2*1=-3

Ответ: -1, -3

Answer 2

x² + 4x + 3 = 0

a = 1; b = 4; c = 3.

D = b² - 4ac= 16 - 12 = 4(>0; 2 корня.)

х1 = (-b - √D) / 2 = (-4 - 2) / 2 = - 3

x2 = (-b - √D) / 2 = ( - 4 + 2) / 2 = - 1

x² + 4x + 3 = 0. (3 + 1 = 4) ( 3 • 1 = 3)

x1 = - 3; x2 = - 1.