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=-(2+sqrt(6))/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=-(2+sqrt(6))/2

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

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

Decimal Notation

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

x^{3} - 2 x^{2} + 3 x - 6

\frac{x + 1}{x + 2} \geq \frac{x - 4}{2 x - 3}

\frac{x + 27}{x + 6} \leq 9

- \frac{4}{- \frac{1}{2}}

3 x - 3 (x + 1) = 3

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=-(2+sqrt(6))/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=-(2+sqrt(6))/2