What are the solutions to the equation -z-1=-3z^2 ?

The solutions to the equation -z-1=-3z^2 are z=-(-1+sqrt(13))/6 ,z=(1+sqrt(13))/6

Find the zeros of -z-1=-3z^2

The zeros of -z-1=-3z^2 are z=-(-1+sqrt(13))/6 ,z=(1+sqrt(13))/6

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

- z - 1 = - 3 z^{2}

z = - \frac{- 1 + 13}{6}, z = \frac{1 + 13}{6}

Decimal Notation

z = - 0.43425 \dots, z = 0.76759 \dots

Solution steps

- z - 1 = - 3 z^{2}

Switch sides

- 3 z^{2} = - z - 1

Move

1

to the left side

- 3 z^{2} + 1 = - z

Move

z

to the left side

- 3 z^{2} + 1 + z = 0

Write in the standard form

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

- 3 z^{2} + z + 1 = 0

Solve with the quadratic formula

z_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 ( - 3 ) \cdot 1}{2 ( - 3 )}

1^{2} - 4 (- 3) \cdot 1 = 13

z_{1, 2} = \frac{- 1 \pm 13}{2 ( - 3 )}

Separate the solutions

z_{1} = \frac{- 1 + 13}{2 ( - 3 )}, z_{2} = \frac{- 1 - 13}{2 ( - 3 )}

z = \frac{- 1 + 13}{2 ( - 3 )} : - \frac{- 1 + 13}{6}

z = \frac{- 1 - 13}{2 ( - 3 )} : \frac{1 + 13}{6}

The solutions to the quadratic equation are:

z = - \frac{- 1 + 13}{6}, z = \frac{1 + 13}{6}

x^{2} - x + 8 = 0

x^{\frac{5}{2}} - 81 x^{\frac{1}{2}}

\frac{2}{3} + \frac{a}{7}

- 7 y = - 77

6 x = - 12

What are the solutions to the equation -z-1=-3z^2 ?
The solutions to the equation -z-1=-3z^2 are z=-(-1+sqrt(13))/6 ,z=(1+sqrt(13))/6
Find the zeros of -z-1=-3z^2
The zeros of -z-1=-3z^2 are z=-(-1+sqrt(13))/6 ,z=(1+sqrt(13))/6