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

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

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

The zeros of quadraticformula 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

(x + 3) (x - 3) = 27

30 + 40 x - 3 x^{2} = 0

\frac{1}{d ^{2}} = \frac{h ^{2} + k ^{2}}{a ^{2}} + \frac{l ^{2}}{c ^{2}}

8 x^{2} + 16 x + 3 = 0

x, x^{2} + y^{2} - 4 x - 2 y + 85 = 0

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