What are the solutions to the equation -2x^2-4x+1=0 ?

The solutions to the equation -2x^2-4x+1=0 are x=-(2+sqrt(6))/2 ,x=(sqrt(6)-2)/2

Find the zeros of -2x^2-4x+1=0

The zeros of -2x^2-4x+1=0 are x=-(2+sqrt(6))/2 ,x=(sqrt(6)-2)/2

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- 2 x^{2} - 4 x + 1 = 0

x = - \frac{2 + 6}{2}, x = \frac{6 - 2}{2}

Solution steps

- 2 x^{2} - 4 x + 1 = 0

Solve with the quadratic formula

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

(- 4)^{2} - 4 (- 2) \cdot 1 = 26

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = - \frac{2 + 6}{2}, x = \frac{6 - 2}{2}

x^{2} + 2 x + 4 = 2 x

7 x^{2} - 105 = 0

2 (3 x^{2} - 1) = - x

- 2 n^{2} = 14

2 x^{2} - 3 = 19

What are the solutions to the equation -2x^2-4x+1=0 ?
The solutions to the equation -2x^2-4x+1=0 are x=-(2+sqrt(6))/2 ,x=(sqrt(6)-2)/2
Find the zeros of -2x^2-4x+1=0
The zeros of -2x^2-4x+1=0 are x=-(2+sqrt(6))/2 ,x=(sqrt(6)-2)/2