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

The solutions to the equation 2x^2-2x-1=0 are x=(1+sqrt(3))/2 ,x=(1-sqrt(3))/2

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

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

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

x = \frac{1 + 3}{2}, x = \frac{1 - 3}{2}

Decimal Notation

x = 1.36602 \dots, x = - 0.36602 \dots

Solution steps

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

Solve with the quadratic formula

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

(- 2)^{2} - 4 \cdot 2 (- 1) = 23

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

Separate the solutions

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

x = \frac{- ( - 2 ) + 2 3}{2 \cdot 2} : \frac{1 + 3}{2}

x = \frac{- ( - 2 ) - 2 3}{2 \cdot 2} : \frac{1 - 3}{2}

The solutions to the quadratic equation are:

x = \frac{1 + 3}{2}, x = \frac{1 - 3}{2}

9 (x - 4) = 18

- 4 (u + 1) = 2 (1 - 2 u) - 6

(x^{2} + 16) (x - 1) (x + 2) \geq 0

b^{2} - 12 b + 36 = 0

(x + 1) (2 x + 5) = (2 x + 3) (x - 4) + 5

What are the solutions to the equation 2x^2-2x-1=0 ?
The solutions to the equation 2x^2-2x-1=0 are x=(1+sqrt(3))/2 ,x=(1-sqrt(3))/2
Find the zeros of 2x^2-2x-1=0
The zeros of 2x^2-2x-1=0 are x=(1+sqrt(3))/2 ,x=(1-sqrt(3))/2