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

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

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

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

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- x^{2} + 2 x = 0

x = 0, x = 2

Solution steps

- x^{2} + 2 x = 0

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = 0, x = 2

\frac{1}{2.5 \cdot 1 0 ^{9}}

- x^{2} + 5 x + 6

7 \cdot 2

5 (4 u + 6) \leq 18 u + 16

\frac{1}{12} (4 x - 3) = \frac{1}{4} (2 x + 1)

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