What are the solutions to the equation 6x^2=-12x-4 ?

The solutions to the equation 6x^2=-12x-4 are x=(-3+sqrt(3))/3 ,x=-(sqrt(3)+1)/(sqrt(3))

Find the zeros of 6x^2=-12x-4

The zeros of 6x^2=-12x-4 are x=(-3+sqrt(3))/3 ,x=-(sqrt(3)+1)/(sqrt(3))

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6 x^{2} = - 12 x - 4

x = \frac{- 3 + 3}{3}, x = - \frac{3 + 1}{3}

Decimal Notation

x = - 0.42264 \dots, x = - 1.57735 \dots

Solution steps

6 x^{2} = - 12 x - 4

Move

4

to the left side

6 x^{2} + 4 = - 12 x

Move

12 x

to the left side

6 x^{2} + 4 + 12 x = 0

Write in the standard form

a x^{2} + b x + c = 0

6 x^{2} + 12 x + 4 = 0

Solve with the quadratic formula

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

1 2^{2} - 4 \cdot 6 \cdot 4 = 43

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

Separate the solutions

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

x = \frac{- 12 + 4 3}{2 \cdot 6} : \frac{- 3 + 3}{3}

x = \frac{- 12 - 4 3}{2 \cdot 6} : - \frac{3 + 1}{3}

The solutions to the quadratic equation are:

x = \frac{- 3 + 3}{3}, x = - \frac{3 + 1}{3}

\frac{8}{25}

\frac{2 a ^{2} + 2 ab}{3 a ^{2} b}

2 (x + 7) \leq 16

21 x^{2} + 14 x

\frac{3}{8} + \frac{1}{5} - \frac{3}{4}

What are the solutions to the equation 6x^2=-12x-4 ?
The solutions to the equation 6x^2=-12x-4 are x=(-3+sqrt(3))/3 ,x=-(sqrt(3)+1)/(sqrt(3))
Find the zeros of 6x^2=-12x-4
The zeros of 6x^2=-12x-4 are x=(-3+sqrt(3))/3 ,x=-(sqrt(3)+1)/(sqrt(3))