What are the solutions to the equation 2x^2+4=2x ?

The solutions to the equation 2x^2+4=2x are x= 1/2+i(sqrt(7))/2 ,x= 1/2-i(sqrt(7))/2

Find the zeros of 2x^2+4=2x

The zeros of 2x^2+4=2x are x= 1/2+i(sqrt(7))/2 ,x= 1/2-i(sqrt(7))/2

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2 x^{2} + 4 = 2 x

x = \frac{1}{2} + i \frac{7}{2}, x = \frac{1}{2} - i \frac{7}{2}

Solution steps

2 x^{2} + 4 = 2 x

Move

2 x

to the left side

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Simplify

(- 2)^{2} - 4 \cdot 2 \cdot 4 : 27 i

x_{1, 2} = \frac{- ( - 2 ) \pm 2 7 i}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 2 ) + 2 7 i}{2 \cdot 2}, x_{2} = \frac{- ( - 2 ) - 2 7 i}{2 \cdot 2}

x = \frac{- ( - 2 ) + 2 7 i}{2 \cdot 2} : \frac{1}{2} + i \frac{7}{2}

x = \frac{- ( - 2 ) - 2 7 i}{2 \cdot 2} : \frac{1}{2} - i \frac{7}{2}

The solutions to the quadratic equation are:

x = \frac{1}{2} + i \frac{7}{2}, x = \frac{1}{2} - i \frac{7}{2}

2 y^{2} + 8 = 0

x^{2} + 50 x = 0

(2 x - 3)^{2} - (x + 5)^{2} = 23

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

(2 a)^{2} + a^{2} + (a - 1)^{2} = (a + 1)^{2} + (a - 2)^{2}

What are the solutions to the equation 2x^2+4=2x ?
The solutions to the equation 2x^2+4=2x are x= 1/2+i(sqrt(7))/2 ,x= 1/2-i(sqrt(7))/2
Find the zeros of 2x^2+4=2x
The zeros of 2x^2+4=2x are x= 1/2+i(sqrt(7))/2 ,x= 1/2-i(sqrt(7))/2