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

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

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

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

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

x = 2

Solution steps

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

Switch sides

1 - 2 x + x^{2} = 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

1872

2 (5 x - 2) = 10 x + 4

\frac{x ^{2}}{4} - 5 x + 17 = 0

\frac{2}{5} - 2

x - 8 > 5 x + 3

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