Question

Помогите пожалуйста!
9x^2+6x+1=0

Answer 1

9x^2+6x+1=0
(3x+1)^2=0 <=> 3x+1=0
x=-1/3

Answer 2

$9x {}^{2} + 6x + 1 = 0 \\ \\ x1.2 = \frac{ - 6 + - \sqrt{6 {}^{2} - 4 \times 9 \times 1 } }{2 \times 9} = \frac{ - 6 + - \sqrt{36 - 36} }{18} = \frac{ - 6 + - \sqrt{0} }{18} = \frac{ - 6 + - 0}{18} \\ \\ x1 = \frac{ - 6 + 0}{18} = \frac{ - 6}{18} = - \frac{1}{3} \\ \\ x2 = \frac{ - 6 - 0}{18} = \frac{ - 6}{18} = - \frac{1}{3}$

Ответ: $- \frac{1}{3}$