What are the solutions to the equation -6y^2-9y=-1 ?

The solutions to the equation -6y^2-9y=-1 are y=-(9+sqrt(105))/(12),y=(sqrt(105)-9)/(12)

Find the zeros of -6y^2-9y=-1

The zeros of -6y^2-9y=-1 are y=-(9+sqrt(105))/(12),y=(sqrt(105)-9)/(12)

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- 6 y^{2} - 9 y = - 1

y = - \frac{9 + 105}{12}, y = \frac{105 - 9}{12}

Solution steps

- 6 y^{2} - 9 y = - 1

Move

1

to the left side

- 6 y^{2} - 9 y + 1 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 9 ) \pm ( - 9 ) ^{2} - 4 ( - 6 ) \cdot 1}{2 ( - 6 )}

(- 9)^{2} - 4 (- 6) \cdot 1 = 105

y_{1, 2} = \frac{- ( - 9 ) \pm 105}{2 ( - 6 )}

Separate the solutions

y_{1} = \frac{- ( - 9 ) + 105}{2 ( - 6 )}, y_{2} = \frac{- ( - 9 ) - 105}{2 ( - 6 )}

y = \frac{- ( - 9 ) + 105}{2 ( - 6 )} : - \frac{9 + 105}{12}

y = \frac{- ( - 9 ) - 105}{2 ( - 6 )} : \frac{105 - 9}{12}

The solutions to the quadratic equation are:

y = - \frac{9 + 105}{12}, y = \frac{105 - 9}{12}

2 y - 1 \leq 7

14 y - y x - 7 y^{2} + 2 x

3 (x - 4)^{2} + 1 = 13

\frac{5 x}{3} + \frac{3 x}{5} = \frac{4}{15}

(\frac{2 r ^{3} t ^{6}}{5 u ^{9}})^{4}

What are the solutions to the equation -6y^2-9y=-1 ?
The solutions to the equation -6y^2-9y=-1 are y=-(9+sqrt(105))/(12),y=(sqrt(105)-9)/(12)
Find the zeros of -6y^2-9y=-1
The zeros of -6y^2-9y=-1 are y=-(9+sqrt(105))/(12),y=(sqrt(105)-9)/(12)