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

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

Find the zeros of 2x^2+(-2)x=-2

The zeros of 2x^2+(-2)x=-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 = - 2

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

Solution steps

2 x^{2} + (- 2) x = - 2

Refine

2 x^{2} - 2 x = - 2

Move

2

to the left side

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}

z^{2} - 2 z - 35 = 0

- 3 x^{2} + 5 x - 7 = 0

\frac{1}{2} \cdot \frac{1}{32} (10 - x) (10 - x) = 0.3

- 32 x^{2} + 19 x + 17 = 0

x^{2} + x + 1.5 = 0

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