What are the solutions to the equation n^2-2n+4=0 ?

The solutions to the equation n^2-2n+4=0 are n=1+sqrt(3)i,n=1-sqrt(3)i

Find the zeros of n^2-2n+4=0

The zeros of n^2-2n+4=0 are n=1+sqrt(3)i,n=1-sqrt(3)i

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n^{2} - 2 n + 4 = 0

n = 1 + 3 i, n = 1 - 3 i

Solution steps

n^{2} - 2 n + 4 = 0

Solve with the quadratic formula

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

Simplify

(- 2)^{2} - 4 \cdot 1 \cdot 4 : 23 i

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

n = 1 + 3 i, n = 1 - 3 i

x^{2} - 4

15 + 3 x^{2} = 24

x^{2} - 2 x + 4

x^{2} - 2 x - 3

3 (x + 5)^{2} = 12

What are the solutions to the equation n^2-2n+4=0 ?
The solutions to the equation n^2-2n+4=0 are n=1+sqrt(3)i,n=1-sqrt(3)i
Find the zeros of n^2-2n+4=0
The zeros of n^2-2n+4=0 are n=1+sqrt(3)i,n=1-sqrt(3)i