What are the solutions to the equation -3z^2-4z+1=0 ?

The solutions to the equation -3z^2-4z+1=0 are z=-(2+sqrt(7))/3 ,z=(sqrt(7)-2)/3

Find the zeros of -3z^2-4z+1=0

The zeros of -3z^2-4z+1=0 are z=-(2+sqrt(7))/3 ,z=(sqrt(7)-2)/3

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- 3 z^{2} - 4 z + 1 = 0

z = - \frac{2 + 7}{3}, z = \frac{7 - 2}{3}

Decimal Notation

z = - 1.54858 \dots, z = 0.21525 \dots

Solution steps

- 3 z^{2} - 4 z + 1 = 0

Solve with the quadratic formula

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

(- 4)^{2} - 4 (- 3) \cdot 1 = 27

z_{1, 2} = \frac{- ( - 4 ) \pm 2 7}{2 ( - 3 )}

Separate the solutions

z_{1} = \frac{- ( - 4 ) + 2 7}{2 ( - 3 )}, z_{2} = \frac{- ( - 4 ) - 2 7}{2 ( - 3 )}

z = \frac{- ( - 4 ) + 2 7}{2 ( - 3 )} : - \frac{2 + 7}{3}

z = \frac{- ( - 4 ) - 2 7}{2 ( - 3 )} : \frac{7 - 2}{3}

The solutions to the quadratic equation are:

z = - \frac{2 + 7}{3}, z = \frac{7 - 2}{3}

\frac{3 ( 4 - z )}{2} = - z

e^{x} \cdot e^{y}

\frac{8}{3} s = - 9 + 7

y^{2} - 8 y + 15

lo g_{6} (1296) = x

What are the solutions to the equation -3z^2-4z+1=0 ?
The solutions to the equation -3z^2-4z+1=0 are z=-(2+sqrt(7))/3 ,z=(sqrt(7)-2)/3
Find the zeros of -3z^2-4z+1=0
The zeros of -3z^2-4z+1=0 are z=-(2+sqrt(7))/3 ,z=(sqrt(7)-2)/3