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

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

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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

Solution steps

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

Write in the standard form

a x^{2} + b x + c = 0

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

(- 2 + i) (- 2 - i)

6 1 1^{5}

(2 k + 3)^{2} = 27

1 6^{2} + x^{2} = 2 0^{2}

∣ 2 x ∣ = 10

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