What are the solutions to the equation n^2-3n+7=0 ?

The solutions to the equation n^2-3n+7=0 are n= 3/2+i(sqrt(19))/2 ,n= 3/2-i(sqrt(19))/2

Find the zeros of n^2-3n+7=0

The zeros of n^2-3n+7=0 are n= 3/2+i(sqrt(19))/2 ,n= 3/2-i(sqrt(19))/2

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n^{2} - 3 n + 7 = 0

n = \frac{3}{2} + i \frac{19}{2}, n = \frac{3}{2} - i \frac{19}{2}

Solution steps

n^{2} - 3 n + 7 = 0

Solve with the quadratic formula

n_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 1 \cdot 7}{2 \cdot 1}

Simplify

(- 3)^{2} - 4 \cdot 1 \cdot 7 : 19 i

n_{1, 2} = \frac{- ( - 3 ) \pm 19 i}{2 \cdot 1}

Separate the solutions

n_{1} = \frac{- ( - 3 ) + 19 i}{2 \cdot 1}, n_{2} = \frac{- ( - 3 ) - 19 i}{2 \cdot 1}

n = \frac{- ( - 3 ) + 19 i}{2 \cdot 1} : \frac{3}{2} + i \frac{19}{2}

n = \frac{- ( - 3 ) - 19 i}{2 \cdot 1} : \frac{3}{2} - i \frac{19}{2}

The solutions to the quadratic equation are:

n = \frac{3}{2} + i \frac{19}{2}, n = \frac{3}{2} - i \frac{19}{2}

x^{2} + 2 x + c

x^{2} - 9 = 3 x + 1

16 - 25 y^{2} = 0

3 x^{2} + 6 x - 29 = - 4

y, x^{2} y = y - 7

What are the solutions to the equation n^2-3n+7=0 ?
The solutions to the equation n^2-3n+7=0 are n= 3/2+i(sqrt(19))/2 ,n= 3/2-i(sqrt(19))/2
Find the zeros of n^2-3n+7=0
The zeros of n^2-3n+7=0 are n= 3/2+i(sqrt(19))/2 ,n= 3/2-i(sqrt(19))/2