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

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

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

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

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

x = - 2

Solution steps

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

Move

12

to the left side

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

- 2 x + 4 = - 3 x + 4

0.21 x + 0.05 (17 - x) \geq 2.55

8 x^{3} - 27 v^{3}

9 x^{3} + 9 x^{2} y - x - y

x^{2} - 12 x + 16 = 0

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