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

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

Find the zeros of 2y^2-4y-1=0

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

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

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

Decimal Notation

y = 2.22474 \dots, y = - 0.22474 \dots

Solution steps

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

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

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

25 x^{2} = 4

363 π - π h^{2} = 0

13 x^{2} = \frac{2}{1 3}

11 b^{2} - 6 b + 1 = 2 b^{2}

2 w^{2} + 4 w = - 3

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