What are the solutions to the equation -4y^2-9y-3=0 ?

The solutions to the equation -4y^2-9y-3=0 are y=-(9+sqrt(33))/8 ,y=-(9-sqrt(33))/8

Find the zeros of -4y^2-9y-3=0

The zeros of -4y^2-9y-3=0 are y=-(9+sqrt(33))/8 ,y=-(9-sqrt(33))/8

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- 4 y^{2} - 9 y - 3 = 0

y = - \frac{9 + 33}{8}, y = - \frac{9 - 33}{8}

Decimal Notation

y = - 1.84307 \dots, y = - 0.40692 \dots

Solution steps

- 4 y^{2} - 9 y - 3 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 9 ) \pm ( - 9 ) ^{2} - 4 ( - 4 ) ( - 3 )}{2 ( - 4 )}

(- 9)^{2} - 4 (- 4) (- 3) = 33

y_{1, 2} = \frac{- ( - 9 ) \pm 33}{2 ( - 4 )}

Separate the solutions

y_{1} = \frac{- ( - 9 ) + 33}{2 ( - 4 )}, y_{2} = \frac{- ( - 9 ) - 33}{2 ( - 4 )}

y = \frac{- ( - 9 ) + 33}{2 ( - 4 )} : - \frac{9 + 33}{8}

y = \frac{- ( - 9 ) - 33}{2 ( - 4 )} : - \frac{9 - 33}{8}

The solutions to the quadratic equation are:

y = - \frac{9 + 33}{8}, y = - \frac{9 - 33}{8}

x^{2} + 15 x = x + 98

x^{2} + 3 x + 5 = 0

x^{2} - 11 x - 42 = 0

- x^{2} = x^{2} - 8

x, e^{t} = x^{2}

What are the solutions to the equation -4y^2-9y-3=0 ?
The solutions to the equation -4y^2-9y-3=0 are y=-(9+sqrt(33))/8 ,y=-(9-sqrt(33))/8
Find the zeros of -4y^2-9y-3=0
The zeros of -4y^2-9y-3=0 are y=-(9+sqrt(33))/8 ,y=-(9-sqrt(33))/8