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

4 x^{2} = 60

y^{2} - 16 y + 64

ln (\frac{4}{5})

x^{2} - 20 = 5 x

5 - x \leq 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