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

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

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

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

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

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

Decimal Notation

x = 2.22474 \dots, x = - 0.22474 \dots

Solution steps

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

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

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

\frac{x ^{2} + 10}{2} = 19

x^{2} - 10 x - 4

2 x^{2} + 5 x + 3

x, x^{2} + x y + 2 y^{2} = 18

x^{2} + 8 = 20

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