What are the solutions to the equation -3z^2-2z+6=0 ?

The solutions to the equation -3z^2-2z+6=0 are z=-(1+sqrt(19))/3 ,z=(sqrt(19)-1)/3

Find the zeros of -3z^2-2z+6=0

The zeros of -3z^2-2z+6=0 are z=-(1+sqrt(19))/3 ,z=(sqrt(19)-1)/3

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- 3 z^{2} - 2 z + 6 = 0

z = - \frac{1 + 19}{3}, z = \frac{19 - 1}{3}

Decimal Notation

z = - 1.78629 \dots, z = 1.11963 \dots

Solution steps

- 3 z^{2} - 2 z + 6 = 0

Solve with the quadratic formula

z_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 ( - 3 ) \cdot 6}{2 ( - 3 )}

(- 2)^{2} - 4 (- 3) \cdot 6 = 219

z_{1, 2} = \frac{- ( - 2 ) \pm 2 19}{2 ( - 3 )}

Separate the solutions

z_{1} = \frac{- ( - 2 ) + 2 19}{2 ( - 3 )}, z_{2} = \frac{- ( - 2 ) - 2 19}{2 ( - 3 )}

z = \frac{- ( - 2 ) + 2 19}{2 ( - 3 )} : - \frac{1 + 19}{3}

z = \frac{- ( - 2 ) - 2 19}{2 ( - 3 )} : \frac{19 - 1}{3}

The solutions to the quadratic equation are:

z = - \frac{1 + 19}{3}, z = \frac{19 - 1}{3}

\frac{x ^{- 1} + y ^{- 1}}{( x + y ) ^{- 1}}

- 6 x^{2} - 15 x + 9

\frac{- \frac{π}{3} + \frac{2 π}{3}}{2}

\frac{1}{x ^{2} + x} \geq \frac{1}{x ^{2} - x} - \frac{1}{x ^{2} - x}

350 \leq 5.5 t - 110

What are the solutions to the equation -3z^2-2z+6=0 ?
The solutions to the equation -3z^2-2z+6=0 are z=-(1+sqrt(19))/3 ,z=(sqrt(19)-1)/3
Find the zeros of -3z^2-2z+6=0
The zeros of -3z^2-2z+6=0 are z=-(1+sqrt(19))/3 ,z=(sqrt(19)-1)/3