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

The solutions to the equation 2x-x^2=-x are x=0,x=3

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

The zeros of 2x-x^2=-x are x=0,x=3

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2 x - x^{2} = - x

x = 0, x = 3

Solution steps

2 x - x^{2} = - x

Move

x

to the left side

- x^{2} + 3 x = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 ( - 1 ) \cdot 0}{2 ( - 1 )}

3^{2} - 4 (- 1) \cdot 0 = 3

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

Separate the solutions

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

x = \frac{- 3 + 3}{2 ( - 1 )} : 0

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

The solutions to the quadratic equation are:

x = 0, x = 3

(x + 2) (x + 2) = 1

\frac{x}{6} = 7

- x^{2} + 2 x > 0

4 x \geq 5 x - 7

64 - 81 d^{2}

What are the solutions to the equation 2x-x^2=-x ?
The solutions to the equation 2x-x^2=-x are x=0,x=3
Find the zeros of 2x-x^2=-x
The zeros of 2x-x^2=-x are x=0,x=3