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

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

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

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

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

x = 2

Solution steps

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

Switch sides

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

Move

3

to the left side

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

Move

2 x

to the left side

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

Solve with the quadratic formula

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

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

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

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

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

x = 2

The solution to the quadratic equation is:

x = 2

- x^{2} - 8

40 x^{2} + 5 x = 0

x^{2} + 4 x + 2.5 = 0

x^{2} - 17 x = - 72

10 x^{2} + 4 x + 77 = 9

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