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=(sqrt(5)-1)/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=(sqrt(5)-1)/2

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

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

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

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

Separate the solutions

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

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

x = \frac{- ( - 2 ) - 2 5}{2 ( - 2 )} : \frac{5 - 1}{2}

The solutions to the quadratic equation are:

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

3 x^{2} - 16 x + 1 = - 4

t^{2} + 7 t = 0

- x^{2} = 1

x^{2} - 6 x + 9 = 9

7 x^{2} - 18 x - 9 = 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=(sqrt(5)-1)/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=(sqrt(5)-1)/2