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

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

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

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

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

y = - 1, y = 2

Solution steps

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

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

y = \frac{- 1 - 3}{2 ( - 1 )} : 2

The solutions to the quadratic equation are:

y = - 1, y = 2

(2 y)^{2} + y^{2} = 1

\frac{x ^{2}}{8} - \frac{x}{4} = \frac{1}{2}

m^{2} + 8 m = - 16

0 = - 2 x^{2}

3 3^{2} + 4 4^{2} = c^{2}

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