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

The solutions to the equation 2x^2+2x-2=0 are x=(-1+sqrt(5))/2 ,x=-(1+sqrt(5))/2

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

The zeros of 2x^2+2x-2=0 are x=(-1+sqrt(5))/2 ,x=-(1+sqrt(5))/2

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

x = \frac{- 1 + 5}{2}, x = - \frac{1 + 5}{2}

Solution steps

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

Solve with the quadratic formula

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

2^{2} - 4 \cdot 2 (- 2) = 25

x_{1, 2} = \frac{- 2 \pm 2 5}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 2 + 2 5}{2 \cdot 2}, x_{2} = \frac{- 2 - 2 5}{2 \cdot 2}

x = \frac{- 2 + 2 5}{2 \cdot 2} : \frac{- 1 + 5}{2}

x = \frac{- 2 - 2 5}{2 \cdot 2} : - \frac{1 + 5}{2}

The solutions to the quadratic equation are:

x = \frac{- 1 + 5}{2}, x = - \frac{1 + 5}{2}

∣ x ∣ \leq - 2

5 x - [7 - (2 x - 1)] = 3 (x - 5) + 4 (x + 3)

- 2 y < - 8

3 - ∣ 3 - 5 x ∣ = - 2

2 m^{2} + \frac{2}{3} m = 0

What are the solutions to the equation 2x^2+2x-2=0 ?
The solutions to the equation 2x^2+2x-2=0 are x=(-1+sqrt(5))/2 ,x=-(1+sqrt(5))/2
Find the zeros of 2x^2+2x-2=0
The zeros of 2x^2+2x-2=0 are x=(-1+sqrt(5))/2 ,x=-(1+sqrt(5))/2