What are the solutions to the equation 3y^2+7y+3=-y^2 ?

The solutions to the equation 3y^2+7y+3=-y^2 are y=-3/4 ,y=-1

Find the zeros of 3y^2+7y+3=-y^2

The zeros of 3y^2+7y+3=-y^2 are y=-3/4 ,y=-1

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3 y^{2} + 7 y + 3 = - y^{2}

y = - \frac{3}{4}, y = - 1

Decimal Notation

y = - 0.75, y = - 1

Solution steps

3 y^{2} + 7 y + 3 = - y^{2}

Move

y^{2}

to the left side

4 y^{2} + 7 y + 3 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 \cdot 4 \cdot 3}{2 \cdot 4}

7^{2} - 4 \cdot 4 \cdot 3 = 1

y_{1, 2} = \frac{- 7 \pm 1}{2 \cdot 4}

Separate the solutions

y_{1} = \frac{- 7 + 1}{2 \cdot 4}, y_{2} = \frac{- 7 - 1}{2 \cdot 4}

y = \frac{- 7 + 1}{2 \cdot 4} : - \frac{3}{4}

y = \frac{- 7 - 1}{2 \cdot 4} : - 1

The solutions to the quadratic equation are:

y = - \frac{3}{4}, y = - 1

- \frac{1}{2} \cdot 2 + 2

6 p + 5 \leq - 2 p - 5

- 3 \leq a \leq 3

\frac{x ^{- 2} + y ^{- 2}}{x ^{- 2} - y ^{- 2}}

y^{2} + 11 y + 14 = 0

What are the solutions to the equation 3y^2+7y+3=-y^2 ?
The solutions to the equation 3y^2+7y+3=-y^2 are y=-3/4 ,y=-1
Find the zeros of 3y^2+7y+3=-y^2
The zeros of 3y^2+7y+3=-y^2 are y=-3/4 ,y=-1