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

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

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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

Decimal Notation

y = 0.61803 \dots, y = - 1.61803 \dots

Solution steps

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

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

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

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

x^{2} = 64

x^{2} + \frac{3}{2} x - 7 = 0

2 x (x + 2) + x + 2 = 5

9 x^{2} - 53 x + 49 = 0

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