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

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

Find the zeros of 3x^2=1-6x

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

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3 x^{2} = 1 - 6 x

x = \frac{- 3 + 2 3}{3}, x = - \frac{3 + 2}{3}

Decimal Notation

x = 0.15470 \dots, x = - 2.15470 \dots

Solution steps

3 x^{2} = 1 - 6 x

Move

6 x

to the left side

3 x^{2} + 6 x = 1

Move

1

to the left side

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

Solve with the quadratic formula

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

6^{2} - 4 \cdot 3 (- 1) = 43

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

Separate the solutions

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

x = \frac{- 6 + 4 3}{2 \cdot 3} : \frac{- 3 + 2 3}{3}

x = \frac{- 6 - 4 3}{2 \cdot 3} : - \frac{3 + 2}{3}

The solutions to the quadratic equation are:

x = \frac{- 3 + 2 3}{3}, x = - \frac{3 + 2}{3}

16 - 4 (4 - x) = 3 (1 - x)

36 x^{2} - 100

3 x < 18

2 (x - 3) (x + 5) = 0

lo g_{4} (\frac{1}{32})

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