What are the solutions to the equation 3y^2+19y+6=0 ?

The solutions to the equation 3y^2+19y+6=0 are y=-1/3 ,y=-6

Find the zeros of 3y^2+19y+6=0

The zeros of 3y^2+19y+6=0 are y=-1/3 ,y=-6

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3 y^{2} + 19 y + 6 = 0

y = - \frac{1}{3}, y = - 6

Solution steps

3 y^{2} + 19 y + 6 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 19 \pm 1 9 ^{2} - 4 \cdot 3 \cdot 6}{2 \cdot 3}

1 9^{2} - 4 \cdot 3 \cdot 6 = 17

y_{1, 2} = \frac{- 19 \pm 17}{2 \cdot 3}

Separate the solutions

y_{1} = \frac{- 19 + 17}{2 \cdot 3}, y_{2} = \frac{- 19 - 17}{2 \cdot 3}

y = \frac{- 19 + 17}{2 \cdot 3} : - \frac{1}{3}

y = \frac{- 19 - 17}{2 \cdot 3} : - 6

The solutions to the quadratic equation are:

y = - \frac{1}{3}, y = - 6

- 4 n + 3 > - 5

8 m - 20 n

2 x^{2} - x \leq 3

w^{4} - 81

(- \frac{2}{3})^{- 2}

What are the solutions to the equation 3y^2+19y+6=0 ?
The solutions to the equation 3y^2+19y+6=0 are y=-1/3 ,y=-6
Find the zeros of 3y^2+19y+6=0
The zeros of 3y^2+19y+6=0 are y=-1/3 ,y=-6