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

The solutions to the equation 2x^2-2x+3=x^2-3x+2 are x=-1/2+i(sqrt(3))/2 ,x=-1/2-i(sqrt(3))/2

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

The zeros of 2x^2-2x+3=x^2-3x+2 are x=-1/2+i(sqrt(3))/2 ,x=-1/2-i(sqrt(3))/2

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

x = - \frac{1}{2} + i \frac{3}{2}, x = - \frac{1}{2} - i \frac{3}{2}

Solution steps

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

Move

2

to the left side

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

Move

3 x

to the left side

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

Move

x^{2}

to the left side

x^{2} + x + 1 = 0

Solve with the quadratic formula

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

Simplify

1^{2} - 4 \cdot 1 \cdot 1 : 3 i

x_{1, 2} = \frac{- 1 \pm 3 i}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 1 + 3 i}{2 \cdot 1}, x_{2} = \frac{- 1 - 3 i}{2 \cdot 1}

x = \frac{- 1 + 3 i}{2 \cdot 1} : - \frac{1}{2} + i \frac{3}{2}

x = \frac{- 1 - 3 i}{2 \cdot 1} : - \frac{1}{2} - i \frac{3}{2}

The solutions to the quadratic equation are:

x = - \frac{1}{2} + i \frac{3}{2}, x = - \frac{1}{2} - i \frac{3}{2}

x, 2 x^{2} + 2 x y + y^{2} = 2

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

x, x^{2} - 3 x - 28 = 0

6 x^{2} + 9 x = 0

x^{2} + 11 x - 242 = 0

What are the solutions to the equation 2x^2-2x+3=x^2-3x+2 ?
The solutions to the equation 2x^2-2x+3=x^2-3x+2 are x=-1/2+i(sqrt(3))/2 ,x=-1/2-i(sqrt(3))/2
Find the zeros of 2x^2-2x+3=x^2-3x+2
The zeros of 2x^2-2x+3=x^2-3x+2 are x=-1/2+i(sqrt(3))/2 ,x=-1/2-i(sqrt(3))/2