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

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

Find the zeros of x^2-x=2*(x^2-2x)

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

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x^{2} - x = 2 \cdot (x^{2} - 2 x)

x = 0, x = 3

Solution steps

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

Expand

2 (x^{2} - 2 x) : 2 x^{2} - 4 x

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

Move

4 x

to the left side

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

Move

2 x^{2}

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

y + 1 = \frac{1}{2} y^{2} - 3

y^{2} - 18 y + 97 = 0

A = x^{2} + 16 x + 64

x^{2} + 10 = 10

- 5 q + 200 = 0.5 q^{2} + 100

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