What are the solutions to the equation 3x^2+6x+1=0 ?

The solutions to the equation 3x^2+6x+1=0 are x=(-3+sqrt(6))/3 ,x=-(3+sqrt(6))/3

Find the zeros of 3x^2+6x+1=0

The zeros of 3x^2+6x+1=0 are x=(-3+sqrt(6))/3 ,x=-(3+sqrt(6))/3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

3 x^{2} + 6 x + 1 = 0

x = \frac{- 3 + 6}{3}, x = - \frac{3 + 6}{3}

Decimal Notation

x = - 0.18350 \dots, x = - 1.81649 \dots

Solution steps

3 x^{2} + 6 x + 1 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 \cdot 3 \cdot 1}{2 \cdot 3}

6^{2} - 4 \cdot 3 \cdot 1 = 26

x_{1, 2} = \frac{- 6 \pm 2 6}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- 6 + 2 6}{2 \cdot 3}, x_{2} = \frac{- 6 - 2 6}{2 \cdot 3}

x = \frac{- 6 + 2 6}{2 \cdot 3} : \frac{- 3 + 6}{3}

x = \frac{- 6 - 2 6}{2 \cdot 3} : - \frac{3 + 6}{3}

The solutions to the quadratic equation are:

x = \frac{- 3 + 6}{3}, x = - \frac{3 + 6}{3}

\frac{\frac{3}{2}}{2}

a^{6} + b^{9}

2 m - [(m - n) - (m + n)]

- \frac{1}{2}

ln (2 x^{2} - x) < 0

What are the solutions to the equation 3x^2+6x+1=0 ?
The solutions to the equation 3x^2+6x+1=0 are x=(-3+sqrt(6))/3 ,x=-(3+sqrt(6))/3
Find the zeros of 3x^2+6x+1=0
The zeros of 3x^2+6x+1=0 are x=(-3+sqrt(6))/3 ,x=-(3+sqrt(6))/3