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

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

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

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

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

x = \frac{- 3 + 3}{3}, x = - \frac{3 + 1}{3}

Decimal Notation

x = - 0.42264 \dots, x = - 1.57735 \dots

Solution steps

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

Solve with the quadratic formula

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

6^{2} - 4 \cdot 3 \cdot 2 = 23

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

Separate the solutions

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

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

x = \frac{- 6 - 2 3}{2 \cdot 3} : - \frac{3 + 1}{3}

The solutions to the quadratic equation are:

x = \frac{- 3 + 3}{3}, x = - \frac{3 + 1}{3}

(3 - 2)^{2}

t^{2} - 2 t - 2 = 0

2 k^{2} + 9 k = - 7

\frac{1}{4} - \frac{1}{3} x = \frac{1}{3}

\frac{∣ x + 2 ∣}{3} = - 3

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