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}

Decimal Notation

x = 1.61803 \dots, x = - 0.61803 \dots

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}

- \frac{1}{4} (12 x + 8) \leq - 2 x + 11

\frac{1}{4} x + \frac{1}{2} = \frac{3}{4} x

c^{2} + 6 c - 27

\frac{4 - 2 i}{i}

28 (1000) - 0.01 (1000)^{2}

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