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

The solutions to the equation -3y^2-3y+1=0 are y=-(3+sqrt(21))/6 ,y=(sqrt(21)-3)/6

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

The zeros of -3y^2-3y+1=0 are y=-(3+sqrt(21))/6 ,y=(sqrt(21)-3)/6

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

y = - \frac{3 + 21}{6}, y = \frac{21 - 3}{6}

Solution steps

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

Solve with the quadratic formula

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

(- 3)^{2} - 4 (- 3) \cdot 1 = 21

y_{1, 2} = \frac{- ( - 3 ) \pm 21}{2 ( - 3 )}

Separate the solutions

y_{1} = \frac{- ( - 3 ) + 21}{2 ( - 3 )}, y_{2} = \frac{- ( - 3 ) - 21}{2 ( - 3 )}

y = \frac{- ( - 3 ) + 21}{2 ( - 3 )} : - \frac{3 + 21}{6}

y = \frac{- ( - 3 ) - 21}{2 ( - 3 )} : \frac{21 - 3}{6}

The solutions to the quadratic equation are:

y = - \frac{3 + 21}{6}, y = \frac{21 - 3}{6}

2 x + 3 = 17

(\frac{1}{17})^{- 2}

5 x + 17 \geq 37

190 m^{3} + 19

4 n^{2} - 49 = 0

What are the solutions to the equation -3y^2-3y+1=0 ?
The solutions to the equation -3y^2-3y+1=0 are y=-(3+sqrt(21))/6 ,y=(sqrt(21)-3)/6
Find the zeros of -3y^2-3y+1=0
The zeros of -3y^2-3y+1=0 are y=-(3+sqrt(21))/6 ,y=(sqrt(21)-3)/6