What are the solutions to the equation x^2-3x+2=1-x ?

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

Find the zeros of x^2-3x+2=1-x

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

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

x = 1

Solution steps

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

Move

x

to the left side

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

Move

1

to the left side

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

Solve with the quadratic formula

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

(- 2)^{2} - 4 \cdot 1 \cdot 1 = 0

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

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

\frac{- ( - 2 )}{2 \cdot 1} = 1

x = 1

The solution to the quadratic equation is:

x = 1

- 11 \geq - 1 - 2 x

6 x = - 2

8 x^{2} - 26 x - 24

ln (\frac{1}{e}) = x

(x + 2) (x + 3) (x - 1) = (x + 4) (x + 4) (x - 4) - 7

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