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

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

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

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

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

y = 1, y = - 2

Solution steps

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

Solve with the quadratic formula

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

2^{2} - 4 \cdot 2 (- 4) = 6

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

Separate the solutions

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

y = \frac{- 2 + 6}{2 \cdot 2} : 1

y = \frac{- 2 - 6}{2 \cdot 2} : - 2

The solutions to the quadratic equation are:

y = 1, y = - 2

20 x^{2} + 4 x = 0

(0 - 2)^{2} - (y - 1)^{2} = 9

0.25 x^{2} - 350 x - 100000 = 0

1 - 3 x - 2 x^{2} = 0

8^{2} + 6^{2} = X^{2}

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