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(2))/2 ,x=-(sqrt(2)+1)/(sqrt(2))

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

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

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

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

Decimal Notation

x = - 0.29289 \dots, x = - 1.70710 \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 \cdot 1}{2 \cdot 2}

4^{2} - 4 \cdot 2 \cdot 1 = 22

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

Separate the solutions

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

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

x = \frac{- 4 - 2 2}{2 \cdot 2} : - \frac{2 + 1}{2}

The solutions to the quadratic equation are:

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

2^{2^{3}}

(\frac{11}{10})^{2}

18 x^{3} - 9 x^{2} - 2 x + 1 < 0

- 2 x \geq 10

3 \leq x < 6

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