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

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

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

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

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- y^{2} + 2 y = 0

y = 0, y = 2

Solution steps

- y^{2} + 2 y = 0

Solve with the quadratic formula

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

2^{2} - 4 (- 1) \cdot 0 = 2

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

Separate the solutions

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

y = \frac{- 2 + 2}{2 ( - 1 )} : 0

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

The solutions to the quadratic equation are:

y = 0, y = 2

3 x^{2} = - 1

X^{2} - 36 = 0

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

5 x^{2} + 17 x - 12 = 0

7 x^{2} - 16 = 0

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