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

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

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

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

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

y = - 2, y = 1

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 )} : - 2

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

The solutions to the quadratic equation are:

y = - 2, y = 1

(x + 8) (x + 3) = 0

5 x^{2} - 10 x + 17 = 0

0 = x^{2} + 5 x

5 x^{2} - 10 x + 15 = 0

34 - 3 (10 - x) = x^{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=-2,y=1
Find the zeros of -y^2-y+2=0
The zeros of -y^2-y+2=0 are y=-2,y=1