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

The solutions to the equation (2-x)(x+1)-x(x-1)=2 are x=0,x=1

Find the zeros of (2-x)(x+1)-x(x-1)=2

The zeros of (2-x)(x+1)-x(x-1)=2 are x=0,x=1

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(2 - x) (x + 1) - x (x - 1) = 2

x = 0, x = 1

Solution steps

(2 - x) (x + 1) - x (x - 1) = 2

Expand

(2 - x) (x + 1) - x (x - 1) : - 2 x^{2} + 2 x + 2

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

Move

2

to the left side

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

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = 0, x = 1

18 = (- 6 + 3)^{2} + (2 - y)^{2}

(3 n - 2) = \frac{n ( 3 n - 1 )}{2}

u^{2} - 2 u - 2 = 0

x^{2} - 2 x - c = 0

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

What are the solutions to the equation (2-x)(x+1)-x(x-1)=2 ?
The solutions to the equation (2-x)(x+1)-x(x-1)=2 are x=0,x=1
Find the zeros of (2-x)(x+1)-x(x-1)=2
The zeros of (2-x)(x+1)-x(x-1)=2 are x=0,x=1