What are the solutions to the equation -x^2+4x=x^2-2x ?

The solutions to the equation -x^2+4x=x^2-2x are x=0,x=3

Find the zeros of -x^2+4x=x^2-2x

The zeros of -x^2+4x=x^2-2x are x=0,x=3

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- x^{2} + 4 x = x^{2} - 2 x

x = 0, x = 3

Solution steps

- x^{2} + 4 x = x^{2} - 2 x

Move

2 x

to the left side

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

Move

x^{2}

to the left side

- 2 x^{2} + 6 x = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 ( - 2 ) \cdot 0}{2 ( - 2 )}

6^{2} - 4 (- 2) \cdot 0 = 6

x_{1, 2} = \frac{- 6 \pm 6}{2 ( - 2 )}

Separate the solutions

x_{1} = \frac{- 6 + 6}{2 ( - 2 )}, x_{2} = \frac{- 6 - 6}{2 ( - 2 )}

x = \frac{- 6 + 6}{2 ( - 2 )} : 0

x = \frac{- 6 - 6}{2 ( - 2 )} : 3

The solutions to the quadratic equation are:

x = 0, x = 3

\frac{1}{3} x - 3 = - 2 x + 6

3 x^{- 2}

8 - p = - 12 - 3 p

3 x - 12 = 9

\frac{3}{1 + 2 i}

What are the solutions to the equation -x^2+4x=x^2-2x ?
The solutions to the equation -x^2+4x=x^2-2x are x=0,x=3
Find the zeros of -x^2+4x=x^2-2x
The zeros of -x^2+4x=x^2-2x are x=0,x=3