What are the solutions to the equation 2y^2+y-1=0 ?

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

Find the zeros of 2y^2+y-1=0

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

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2 y^{2} + y - 1 = 0

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

Solution steps

2 y^{2} + y - 1 = 0

Solve with the quadratic formula

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

1^{2} - 4 \cdot 2 (- 1) = 3

y_{1, 2} = \frac{- 1 \pm 3}{2 \cdot 2}

Separate the solutions

y_{1} = \frac{- 1 + 3}{2 \cdot 2}, y_{2} = \frac{- 1 - 3}{2 \cdot 2}

y = \frac{- 1 + 3}{2 \cdot 2} : \frac{1}{2}

y = \frac{- 1 - 3}{2 \cdot 2} : - 1

The solutions to the quadratic equation are:

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

x^{2} + 10 x + 23 = 0

28 v^{3} + 16 v^{2} - 21 v - 12

4 (5 x + 2) \leq 28

- 3 \geq - \frac{1}{3} t

\frac{8 a + 2}{5} - 1 = \frac{10 a - 3}{5}

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