What are the solutions to the equation 2y^2+6y=-2 ?

The solutions to the equation 2y^2+6y=-2 are y=(-3+sqrt(5))/2 ,y=-(3+sqrt(5))/2

Find the zeros of 2y^2+6y=-2

The zeros of 2y^2+6y=-2 are y=(-3+sqrt(5))/2 ,y=-(3+sqrt(5))/2

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2 y^{2} + 6 y = - 2

y = \frac{- 3 + 5}{2}, y = - \frac{3 + 5}{2}

Decimal Notation

y = - 0.38196 \dots, y = - 2.61803 \dots

Solution steps

2 y^{2} + 6 y = - 2

Move

2

to the left side

2 y^{2} + 6 y + 2 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 \cdot 2 \cdot 2}{2 \cdot 2}

6^{2} - 4 \cdot 2 \cdot 2 = 25

y_{1, 2} = \frac{- 6 \pm 2 5}{2 \cdot 2}

Separate the solutions

y_{1} = \frac{- 6 + 2 5}{2 \cdot 2}, y_{2} = \frac{- 6 - 2 5}{2 \cdot 2}

y = \frac{- 6 + 2 5}{2 \cdot 2} : \frac{- 3 + 5}{2}

y = \frac{- 6 - 2 5}{2 \cdot 2} : - \frac{3 + 5}{2}

The solutions to the quadratic equation are:

y = \frac{- 3 + 5}{2}, y = - \frac{3 + 5}{2}

6 x^{2} + 12 - 90

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

x^{2} = 6 x + 16

w + \frac{1}{6} = - \frac{2}{3}

(3 x - 2) (2 x + 7) - (3 x + 11) (2 x - 3) = 0

What are the solutions to the equation 2y^2+6y=-2 ?
The solutions to the equation 2y^2+6y=-2 are y=(-3+sqrt(5))/2 ,y=-(3+sqrt(5))/2
Find the zeros of 2y^2+6y=-2
The zeros of 2y^2+6y=-2 are y=(-3+sqrt(5))/2 ,y=-(3+sqrt(5))/2