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

The solutions to the equation -2x+2x^2=-3 are x= 1/2+i(sqrt(5))/2 ,x= 1/2-i(sqrt(5))/2

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

The zeros of -2x+2x^2=-3 are x= 1/2+i(sqrt(5))/2 ,x= 1/2-i(sqrt(5))/2

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

x = \frac{1}{2} + i \frac{5}{2}, x = \frac{1}{2} - i \frac{5}{2}

Solution steps

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

Move

3

to the left side

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

Write in the standard form

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

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

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 2 \cdot 3}{2 \cdot 2}

Simplify

(- 2)^{2} - 4 \cdot 2 \cdot 3 : 25 i

x_{1, 2} = \frac{- ( - 2 ) \pm 2 5 i}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 2 ) + 2 5 i}{2 \cdot 2}, x_{2} = \frac{- ( - 2 ) - 2 5 i}{2 \cdot 2}

x = \frac{- ( - 2 ) + 2 5 i}{2 \cdot 2} : \frac{1}{2} + i \frac{5}{2}

x = \frac{- ( - 2 ) - 2 5 i}{2 \cdot 2} : \frac{1}{2} - i \frac{5}{2}

The solutions to the quadratic equation are:

x = \frac{1}{2} + i \frac{5}{2}, x = \frac{1}{2} - i \frac{5}{2}

2 b^{2} + 12 b + 23 = 6

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

25 x^{2} - 225 x = 0

12 y^{2} - 9 y - 3 = 0

3 (2 x - 6)^{2} + 6 = 27

What are the solutions to the equation -2x+2x^2=-3 ?
The solutions to the equation -2x+2x^2=-3 are x= 1/2+i(sqrt(5))/2 ,x= 1/2-i(sqrt(5))/2
Find the zeros of -2x+2x^2=-3
The zeros of -2x+2x^2=-3 are x= 1/2+i(sqrt(5))/2 ,x= 1/2-i(sqrt(5))/2