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

The solutions to the equation y^2+9y+3=-4y^2 are y=(-9+sqrt(21))/(10),y=(-9-sqrt(21))/(10)

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

The zeros of y^2+9y+3=-4y^2 are y=(-9+sqrt(21))/(10),y=(-9-sqrt(21))/(10)

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

y = \frac{- 9 + 21}{10}, y = \frac{- 9 - 21}{10}

Decimal Notation

y = - 0.44174 \dots, y = - 1.35825 \dots

Solution steps

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

Move

4 y^{2}

to the left side

5 y^{2} + 9 y + 3 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 9 \pm 9 ^{2} - 4 \cdot 5 \cdot 3}{2 \cdot 5}

9^{2} - 4 \cdot 5 \cdot 3 = 21

y_{1, 2} = \frac{- 9 \pm 21}{2 \cdot 5}

Separate the solutions

y_{1} = \frac{- 9 + 21}{2 \cdot 5}, y_{2} = \frac{- 9 - 21}{2 \cdot 5}

y = \frac{- 9 + 21}{2 \cdot 5} : \frac{- 9 + 21}{10}

y = \frac{- 9 - 21}{2 \cdot 5} : \frac{- 9 - 21}{10}

The solutions to the quadratic equation are:

y = \frac{- 9 + 21}{10}, y = \frac{- 9 - 21}{10}

\frac{1}{4} (x + 1)^{2} - 16 = 0

2 x^{2} - 3 x = 3

x, \frac{x ^{2} + y ^{2}}{z ^{2}} = - w^{2}

2 x^{2} - 14 x = - 9 x + 7

x^{2} - 6 x - 8 = - 1

What are the solutions to the equation y^2+9y+3=-4y^2 ?
The solutions to the equation y^2+9y+3=-4y^2 are y=(-9+sqrt(21))/(10),y=(-9-sqrt(21))/(10)
Find the zeros of y^2+9y+3=-4y^2
The zeros of y^2+9y+3=-4y^2 are y=(-9+sqrt(21))/(10),y=(-9-sqrt(21))/(10)