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

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

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

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

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

x = 0, x = 2

Solution steps

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

Switch sides

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

Move

x^{2}

to the left side

- 2 x^{2} + 4 x = 0

Solve with the quadratic formula

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

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

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

Separate the solutions

x_{1} = \frac{- 4 + 4}{2 ( - 2 )}, x_{2} = \frac{- 4 - 4}{2 ( - 2 )}

x = \frac{- 4 + 4}{2 ( - 2 )} : 0

x = \frac{- 4 - 4}{2 ( - 2 )} : 2

The solutions to the quadratic equation are:

x = 0, x = 2

10 \leq 8 - 2 x \leq 18

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

1.76 = \frac{x - 693366}{693366} \cdot 100

x^{2} - x - \frac{5}{2} = 0

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