What are the solutions to the equation 3x^2+12x=-6 ?

The solutions to the equation 3x^2+12x=-6 are x=-2+sqrt(2),x=-2-sqrt(2)

Find the zeros of 3x^2+12x=-6

The zeros of 3x^2+12x=-6 are x=-2+sqrt(2),x=-2-sqrt(2)

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3 x^{2} + 12 x = - 6

x = - 2 + 2, x = - 2 - 2

Solution steps

3 x^{2} + 12 x = - 6

Move

6

to the left side

3 x^{2} + 12 x + 6 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 \cdot 3 \cdot 6}{2 \cdot 3}

1 2^{2} - 4 \cdot 3 \cdot 6 = 62

x_{1, 2} = \frac{- 12 \pm 6 2}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- 12 + 6 2}{2 \cdot 3}, x_{2} = \frac{- 12 - 6 2}{2 \cdot 3}

x = \frac{- 12 + 6 2}{2 \cdot 3} : - 2 + 2

x = \frac{- 12 - 6 2}{2 \cdot 3} : - 2 - 2

The solutions to the quadratic equation are:

x = - 2 + 2, x = - 2 - 2

4 \geq 6 x - 12 - 8

3 y^{2} - 27 y = 0

(1 + \frac{0.052}{12})^{12} - 1

- 3 = \frac{4}{3} x

\frac{365}{180} \frac{100000 - 95400}{95400}

What are the solutions to the equation 3x^2+12x=-6 ?
The solutions to the equation 3x^2+12x=-6 are x=-2+sqrt(2),x=-2-sqrt(2)
Find the zeros of 3x^2+12x=-6
The zeros of 3x^2+12x=-6 are x=-2+sqrt(2),x=-2-sqrt(2)