Question

Решите уравнение:
в)x^2+3x+x+3=0

Answer 1

$displaystyle x^2+3x+x+3=0\\x^2+4x+3=0\\D=4^2-4*3=16-12=4\\x_1=frac{-4+2}2=-frac{2}2=boxed{-1}\\x_2=frac{-4-2}2=boxed{-3}$