What are the solutions to the equation -3z^2-2z+4=0 ?

The solutions to the equation -3z^2-2z+4=0 are z=-(1+sqrt(13))/3 ,z=(sqrt(13)-1)/3

Find the zeros of -3z^2-2z+4=0

The zeros of -3z^2-2z+4=0 are z=-(1+sqrt(13))/3 ,z=(sqrt(13)-1)/3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

- 3 z^{2} - 2 z + 4 = 0

z = - \frac{1 + 13}{3}, z = \frac{13 - 1}{3}

Decimal Notation

z = - 1.53518 \dots, z = 0.86851 \dots

Solution steps

- 3 z^{2} - 2 z + 4 = 0

Solve with the quadratic formula

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

(- 2)^{2} - 4 (- 3) \cdot 4 = 213

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

z = - \frac{1 + 13}{3}, z = \frac{13 - 1}{3}

5 + 4 x + (- 3) = - 10

- 5 \leq x < 2

\frac{x}{5} = 2 x - 6

x^{- 7}

(x^{9})^{\frac{1}{3}}

What are the solutions to the equation -3z^2-2z+4=0 ?
The solutions to the equation -3z^2-2z+4=0 are z=-(1+sqrt(13))/3 ,z=(sqrt(13)-1)/3
Find the zeros of -3z^2-2z+4=0
The zeros of -3z^2-2z+4=0 are z=-(1+sqrt(13))/3 ,z=(sqrt(13)-1)/3