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+2sqrt(3))/3 ,x=(2sqrt(3)-3)/3

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

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

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

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

Decimal Notation

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

Solution steps

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

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

x = \frac{- ( - 6 ) - 4 3}{2 ( - 3 )} : \frac{2 3 - 3}{3}

The solutions to the quadratic equation are:

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

y^{2} + 2 = 0

- 15 x (x - 8) = 0

x, z = 4 x^{2} + 3 y^{2}

y, x = y^{2} + z^{2}

x^{2} - 4 \cdot x + 3 = 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+2sqrt(3))/3 ,x=(2sqrt(3)-3)/3
Find the zeros of -3x^2-6x+1=0
The zeros of -3x^2-6x+1=0 are x=-(3+2sqrt(3))/3 ,x=(2sqrt(3)-3)/3