What are the solutions to the equation x^2-x+6=0 ?

The solutions to the equation x^2-x+6=0 are x= 1/2+i(sqrt(23))/2 ,x= 1/2-i(sqrt(23))/2

Find the zeros of x^2-x+6=0

The zeros of x^2-x+6=0 are x= 1/2+i(sqrt(23))/2 ,x= 1/2-i(sqrt(23))/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

x^{2} - x + 6 = 0

x = \frac{1}{2} + i \frac{23}{2}, x = \frac{1}{2} - i \frac{23}{2}

Solution steps

x^{2} - x + 6 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 1 \cdot 6}{2 \cdot 1}

Simplify

(- 1)^{2} - 4 \cdot 1 \cdot 6 : 23 i

x_{1, 2} = \frac{- ( - 1 ) \pm 23 i}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 1 ) + 23 i}{2 \cdot 1}, x_{2} = \frac{- ( - 1 ) - 23 i}{2 \cdot 1}

x = \frac{- ( - 1 ) + 23 i}{2 \cdot 1} : \frac{1}{2} + i \frac{23}{2}

x = \frac{- ( - 1 ) - 23 i}{2 \cdot 1} : \frac{1}{2} - i \frac{23}{2}

The solutions to the quadratic equation are:

x = \frac{1}{2} + i \frac{23}{2}, x = \frac{1}{2} - i \frac{23}{2}

\frac{4}{3} π 3^{3}

(- \frac{8}{2})^{2}

\frac{13 !}{( 13 - 6 ) ! 6 !}

\frac{1}{2} ln (3)

\frac{1}{2} \cdot 1.204 \cdot (6.1)^{2}

What are the solutions to the equation x^2-x+6=0 ?
The solutions to the equation x^2-x+6=0 are x= 1/2+i(sqrt(23))/2 ,x= 1/2-i(sqrt(23))/2
Find the zeros of x^2-x+6=0
The zeros of x^2-x+6=0 are x= 1/2+i(sqrt(23))/2 ,x= 1/2-i(sqrt(23))/2