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

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

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

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

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

x = 3, x = 0

Solution steps

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

Move

4 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 \cdot 2 \cdot 0}{2 \cdot 2}

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = 3, x = 0

t, f (t - 3) = - (t - 3)^{2} + 6 (t - 3) - 11

8 x^{2} + 128 = 0

y, x^{2} + y^{2} = 25

x^{2} + 10 x + 25 = 36

x^{2} = 3 x + 1

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