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

The solutions to the equation 6y^2-y+2=0 are y= 1/12+i(sqrt(47))/(12),y= 1/12-i(sqrt(47))/(12)

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

The zeros of 6y^2-y+2=0 are y= 1/12+i(sqrt(47))/(12),y= 1/12-i(sqrt(47))/(12)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

6 y^{2} - y + 2 = 0

y = \frac{1}{12} + i \frac{47}{12}, y = \frac{1}{12} - i \frac{47}{12}

Solution steps

6 y^{2} - y + 2 = 0

Solve with the quadratic formula

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

Simplify

(- 1)^{2} - 4 \cdot 6 \cdot 2 : 47 i

y_{1, 2} = \frac{- ( - 1 ) \pm 47 i}{2 \cdot 6}

Separate the solutions

y_{1} = \frac{- ( - 1 ) + 47 i}{2 \cdot 6}, y_{2} = \frac{- ( - 1 ) - 47 i}{2 \cdot 6}

y = \frac{- ( - 1 ) + 47 i}{2 \cdot 6} : \frac{1}{12} + i \frac{47}{12}

y = \frac{- ( - 1 ) - 47 i}{2 \cdot 6} : \frac{1}{12} - i \frac{47}{12}

The solutions to the quadratic equation are:

y = \frac{1}{12} + i \frac{47}{12}, y = \frac{1}{12} - i \frac{47}{12}

3 x^{2} + 10 x - 88 = 0

(x - 6)^{2} - 64 = 0

12 x^{2} - 4 x + 7 = 0

2 (x + 8)^{2} - 3 = 97

(\frac{3 a}{8})^{2} + a = 52

What are the solutions to the equation 6y^2-y+2=0 ?
The solutions to the equation 6y^2-y+2=0 are y= 1/12+i(sqrt(47))/(12),y= 1/12-i(sqrt(47))/(12)
Find the zeros of 6y^2-y+2=0
The zeros of 6y^2-y+2=0 are y= 1/12+i(sqrt(47))/(12),y= 1/12-i(sqrt(47))/(12)