What are the solutions to the equation 2x^2-2x+7=0 ?

The solutions to the equation 2x^2-2x+7=0 are x= 1/2+i(sqrt(13))/2 ,x= 1/2-i(sqrt(13))/2

Find the zeros of 2x^2-2x+7=0

The zeros of 2x^2-2x+7=0 are x= 1/2+i(sqrt(13))/2 ,x= 1/2-i(sqrt(13))/2

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

x = \frac{1}{2} + i \frac{13}{2}, x = \frac{1}{2} - i \frac{13}{2}

Solution steps

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

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 2 \cdot 7}{2 \cdot 2}

Simplify

(- 2)^{2} - 4 \cdot 2 \cdot 7 : 213 i

x_{1, 2} = \frac{- ( - 2 ) \pm 2 13 i}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 2 ) + 2 13 i}{2 \cdot 2}, x_{2} = \frac{- ( - 2 ) - 2 13 i}{2 \cdot 2}

x = \frac{- ( - 2 ) + 2 13 i}{2 \cdot 2} : \frac{1}{2} + i \frac{13}{2}

x = \frac{- ( - 2 ) - 2 13 i}{2 \cdot 2} : \frac{1}{2} - i \frac{13}{2}

The solutions to the quadratic equation are:

x = \frac{1}{2} + i \frac{13}{2}, x = \frac{1}{2} - i \frac{13}{2}

\frac{3 x - 1}{2 x + 5} < 0

27 x^{3} - 64

4 (b - 2) = 2 (5 - b)

25 m^{4} - 30 m^{2} + 9

8 < 4 + x

What are the solutions to the equation 2x^2-2x+7=0 ?
The solutions to the equation 2x^2-2x+7=0 are x= 1/2+i(sqrt(13))/2 ,x= 1/2-i(sqrt(13))/2
Find the zeros of 2x^2-2x+7=0
The zeros of 2x^2-2x+7=0 are x= 1/2+i(sqrt(13))/2 ,x= 1/2-i(sqrt(13))/2