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

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

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

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

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

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

Solution steps

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

Solve with the quadratic formula

n_{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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

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

(1 - x)^{2} = - 1

4 x^{2} - 32 x + 65 = 0

2 x^{2} + 13 x - 45 = 0

x (x + 8) = 6 x + 7

5 y^{2} + 6 = 16

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