What are the solutions to the equation 2y^2+4y+2=0 ?

The solutions to the equation 2y^2+4y+2=0 are y=-1

Find the zeros of 2y^2+4y+2=0

The zeros of 2y^2+4y+2=0 are y=-1

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2 y^{2} + 4 y + 2 = 0

y = - 1

Solution steps

2 y^{2} + 4 y + 2 = 0

Solve with the quadratic formula

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

4^{2} - 4 \cdot 2 \cdot 2 = 0

y_{1, 2} = \frac{- 4 \pm 0}{2 \cdot 2}

y = \frac{- 4}{2 \cdot 2}

\frac{- 4}{2 \cdot 2} = - 1

y = - 1

The solution to the quadratic equation is:

y = - 1

x, x^{2} y^{2} = 25

x^{2} - \frac{5}{3} x = 0

1 - y^{2} = y^{2} - 1

x^{2} + 16 x - 80 = 0

5 s^{2} + 8 s + 18 = 0

What are the solutions to the equation 2y^2+4y+2=0 ?
The solutions to the equation 2y^2+4y+2=0 are y=-1
Find the zeros of 2y^2+4y+2=0
The zeros of 2y^2+4y+2=0 are y=-1