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

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

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

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

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2 x^{2} + 2 = 2 x

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

Solution steps

2 x^{2} + 2 = 2 x

Move

2 x

to the left side

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

Write in the standard form

a x^{2} + b x + c = 0

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

Solve with the quadratic formula

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

Simplify

(- 2)^{2} - 4 \cdot 2 \cdot 2 : 23 i

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

Separate the solutions

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

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

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

3 (y + 2) - y = 2 (y - 1) + 5

7 x + 21

14 x = 12 x + 3

24 n - 42

∣ 2 x - 3 ∣ \leq \frac{2}{5}

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