What are the solutions to the equation 3y^2+12y+12=0 ?

The solutions to the equation 3y^2+12y+12=0 are y=-2

Find the zeros of 3y^2+12y+12=0

The zeros of 3y^2+12y+12=0 are y=-2

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3 y^{2} + 12 y + 12 = 0

y = - 2

Solution steps

3 y^{2} + 12 y + 12 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 \cdot 3 \cdot 12}{2 \cdot 3}

1 2^{2} - 4 \cdot 3 \cdot 12 = 0

y_{1, 2} = \frac{- 12 \pm 0}{2 \cdot 3}

y = \frac{- 12}{2 \cdot 3}

\frac{- 12}{2 \cdot 3} = - 2

y = - 2

The solution to the quadratic equation is:

y = - 2

18 x^{2} - 12 x = 0

x + x^{2} = 2

2 (x - 1)^{2} = 98

6 + b^{2} = 70

2 1^{2} + 2 8^{2} = x^{2}

What are the solutions to the equation 3y^2+12y+12=0 ?
The solutions to the equation 3y^2+12y+12=0 are y=-2
Find the zeros of 3y^2+12y+12=0
The zeros of 3y^2+12y+12=0 are y=-2