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

The solutions to the equation 2y^2+9y+3=4 are y=(-9+sqrt(89))/4 ,y=(-9-sqrt(89))/4

Find the zeros of 2y^2+9y+3=4

The zeros of 2y^2+9y+3=4 are y=(-9+sqrt(89))/4 ,y=(-9-sqrt(89))/4

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2 y^{2} + 9 y + 3 = 4

y = \frac{- 9 + 89}{4}, y = \frac{- 9 - 89}{4}

Decimal

y = 0.10849 \dots, y = - 4.60849 \dots

Solution steps

2 y^{2} + 9 y + 3 = 4

Move

4

to the left side

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

Solve with the quadratic formula

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

9^{2} - 4 \cdot 2 (- 1) = 89

y_{1, 2} = \frac{- 9 \pm 89}{2 \cdot 2}

Separate the solutions

y_{1} = \frac{- 9 + 89}{2 \cdot 2}, y_{2} = \frac{- 9 - 89}{2 \cdot 2}

y = \frac{- 9 + 89}{2 \cdot 2} : \frac{- 9 + 89}{4}

y = \frac{- 9 - 89}{2 \cdot 2} : \frac{- 9 - 89}{4}

The solutions to the quadratic equation are:

y = \frac{- 9 + 89}{4}, y = \frac{- 9 - 89}{4}

5 n^{2} - 7 n + 3 = 2

- 5 x^{2} = 60

- 2 x^{2} - 22 x - 3 = - 10 x + 13

22 c^{2} - 35 c + 12 = - 3 c^{2}

z^{2} + 12 z = - 11

What are the solutions to the equation 2y^2+9y+3=4 ?
The solutions to the equation 2y^2+9y+3=4 are y=(-9+sqrt(89))/4 ,y=(-9-sqrt(89))/4
Find the zeros of 2y^2+9y+3=4
The zeros of 2y^2+9y+3=4 are y=(-9+sqrt(89))/4 ,y=(-9-sqrt(89))/4