What is the answer to (x-1)^2+(1-x)\times (x+2)=0 ?

The answer to (x-1)^2+(1-x)\times (x+2)=0 is x=1

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(x-1)^2+(1-x)\times (x+2)=0

(x - 1)^{2} + (1 - x) \cdot (x + 2) = 0

x = 1

Solution steps

(x - 1)^{2} + (1 - x) (x + 2) = 0

Expand

(x - 1)^{2} + (1 - x) (x + 2) : - 3 x + 3

- 3 x + 3 = 0

Move

3

to the right side

- 3 x = - 3

Divide both sides by

- 3

x = 1

x^{2} - 7 x = 8

2.25 (\frac{1}{100})

(x + 1)^{2} = 25

3 (v + 2) + v = 4 (v - 1) + 10

\frac{2}{5} t - \frac{6}{25} = \frac{3}{5} t + 3

What is the answer to (x-1)^2+(1-x)\times (x+2)=0 ?
The answer to (x-1)^2+(1-x)\times (x+2)=0 is x=1
Solve for x: (x-1)^2+(1-x)\times (x+2)=0
The solution is x=1