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

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

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

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

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- z^{2} + z = 0

z = 0, z = 1

Solution steps

- z^{2} + z = 0

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

z = 0, z = 1

x^{2} - 6 x + 9 = 80

3 (x - 7)^{2} - 36 = 12

(x + 10)^{2} - 14 = 11

16 - 9 x^{2} = 0

x^{2} + 2 x + 3

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