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

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

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

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

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

x = - 2

Solution steps

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

Switch sides

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

15 x^{2} - 35 x - 30

4 x^{2} - 16 x + 10 = 0

\frac{x ^{3}}{x ^{2}}

\frac{x}{3} - 8 = - 2

∣ 2 x + 5 ∣ \geq ∣ x + 4 ∣

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