What are the solutions to the equation x^2-2x+1-x=-2x^2+x ?

The solutions to the equation x^2-2x+1-x=-2x^2+x are x=1,x= 1/3

Find the zeros of x^2-2x+1-x=-2x^2+x

The zeros of x^2-2x+1-x=-2x^2+x are x=1,x= 1/3

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x^{2} - 2 x + 1 - x = - 2 x^{2} + x

x = 1, x = \frac{1}{3}

Solution steps

x^{2} - 2 x + 1 - x = - 2 x^{2} + x

Refine

x^{2} - 3 x + 1 = - 2 x^{2} + x

Move

x

to the left side

x^{2} - 4 x + 1 = - 2 x^{2}

Move

2 x^{2}

to the left side

3 x^{2} - 4 x + 1 = 0

Solve with the quadratic formula

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

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

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

Separate the solutions

x_{1} = \frac{- ( - 4 ) + 2}{2 \cdot 3}, x_{2} = \frac{- ( - 4 ) - 2}{2 \cdot 3}

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

x = \frac{- ( - 4 ) - 2}{2 \cdot 3} : \frac{1}{3}

The solutions to the quadratic equation are:

x = 1, x = \frac{1}{3}

5 (x - 2)^{2} = - 20

- x^{2} + 7 x + 3 = 0

x (2 x + 3) - 5 = 6 x

x^{2} + 20 x - 2400 = 0

3 (- 3)^{2} + 2 y^{2} = 36

What are the solutions to the equation x^2-2x+1-x=-2x^2+x ?
The solutions to the equation x^2-2x+1-x=-2x^2+x are x=1,x= 1/3
Find the zeros of x^2-2x+1-x=-2x^2+x
The zeros of x^2-2x+1-x=-2x^2+x are x=1,x= 1/3