What are the solutions to the equation (n)=n^2-4n+20 ?

The solutions to the equation (n)=n^2-4n+20 are n= 5/2+i(sqrt(55))/2 ,n= 5/2-i(sqrt(55))/2

Find the zeros of (n)=n^2-4n+20

The zeros of (n)=n^2-4n+20 are n= 5/2+i(sqrt(55))/2 ,n= 5/2-i(sqrt(55))/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(n) = n^{2} - 4 n + 20

n = \frac{5}{2} + i \frac{55}{2}, n = \frac{5}{2} - i \frac{55}{2}

Solution steps

(n) = n^{2} - 4 n + 20

Refine

n = n^{2} - 4 n + 20

Switch sides

n^{2} - 4 n + 20 = n

Move

n

to the left side

n^{2} - 5 n + 20 = 0

Solve with the quadratic formula

n_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 1 \cdot 20}{2 \cdot 1}

Simplify

(- 5)^{2} - 4 \cdot 1 \cdot 20 : 55 i

n_{1, 2} = \frac{- ( - 5 ) \pm 55 i}{2 \cdot 1}

Separate the solutions

n_{1} = \frac{- ( - 5 ) + 55 i}{2 \cdot 1}, n_{2} = \frac{- ( - 5 ) - 55 i}{2 \cdot 1}

n = \frac{- ( - 5 ) + 55 i}{2 \cdot 1} : \frac{5}{2} + i \frac{55}{2}

n = \frac{- ( - 5 ) - 55 i}{2 \cdot 1} : \frac{5}{2} - i \frac{55}{2}

The solutions to the quadratic equation are:

n = \frac{5}{2} + i \frac{55}{2}, n = \frac{5}{2} - i \frac{55}{2}

- x^{2} + 7 x = 11.25

w^{2} - 10 w

r^{2} + r - 4 = 0

(h + 1)^{2} = 16

x^{2} + 11 x = 180

What are the solutions to the equation (n)=n^2-4n+20 ?
The solutions to the equation (n)=n^2-4n+20 are n= 5/2+i(sqrt(55))/2 ,n= 5/2-i(sqrt(55))/2
Find the zeros of (n)=n^2-4n+20
The zeros of (n)=n^2-4n+20 are n= 5/2+i(sqrt(55))/2 ,n= 5/2-i(sqrt(55))/2