What are the solutions to the equation 3y^2+28y+9=0 ?

The solutions to the equation 3y^2+28y+9=0 are y=-1/3 ,y=-9

Find the zeros of 3y^2+28y+9=0

The zeros of 3y^2+28y+9=0 are y=-1/3 ,y=-9

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3 y^{2} + 28 y + 9 = 0

y = - \frac{1}{3}, y = - 9

Solution steps

3 y^{2} + 28 y + 9 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 28 \pm 2 8 ^{2} - 4 \cdot 3 \cdot 9}{2 \cdot 3}

2 8^{2} - 4 \cdot 3 \cdot 9 = 26

y_{1, 2} = \frac{- 28 \pm 26}{2 \cdot 3}

Separate the solutions

y_{1} = \frac{- 28 + 26}{2 \cdot 3}, y_{2} = \frac{- 28 - 26}{2 \cdot 3}

y = \frac{- 28 + 26}{2 \cdot 3} : - \frac{1}{3}

y = \frac{- 28 - 26}{2 \cdot 3} : - 9

The solutions to the quadratic equation are:

y = - \frac{1}{3}, y = - 9

2 x^{2} + 50 = 0

4 x^{2} + 9 x + 2 < 0

- x^{2} - 2 x + 1 = 0

ln (x - 2) - 3 ln (2) = 2

x^{2} - 1 \leq ∣ x + 1 ∣

What are the solutions to the equation 3y^2+28y+9=0 ?
The solutions to the equation 3y^2+28y+9=0 are y=-1/3 ,y=-9
Find the zeros of 3y^2+28y+9=0
The zeros of 3y^2+28y+9=0 are y=-1/3 ,y=-9