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

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

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

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

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

x = - 2

Solution steps

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

Solve with the quadratic formula

x_{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

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

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

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

x = - 2

The solution to the quadratic equation is:

x = - 2

12 = - 7 f - 9

2 x^{4} + 3 x^{3} + 48 x^{2} + 75 x - 50

1 5^{2} + 1 0^{2} = c^{2}

∣ 4 x - 3 ∣ \leq 12

x - 3 = 4

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