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

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

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

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

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

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

Decimal Notation

s = - 0.42264 \dots, s = - 1.57735 \dots

Solution steps

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

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

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

4 5^{2} + 6 0^{2} = x^{2}

- \frac{2}{3} (x - 1) (x + 3) = 0

- y^{2} - 4 y - 4 = 0

12 x - x^{2} = 36

2 x^{2} + 8 x + 15 = 0

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