What are the solutions to the equation 2(3x^2-1)=-x ?

The solutions to the equation 2(3x^2-1)=-x are x= 1/2 ,x=-2/3

Find the zeros of 2(3x^2-1)=-x

The zeros of 2(3x^2-1)=-x are x= 1/2 ,x=-2/3

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2 (3 x^{2} - 1) = - x

x = \frac{1}{2}, x = - \frac{2}{3}

Decimal Notation

x = 0.5, x = - 0.66666 \dots

Solution steps

2 (3 x^{2} - 1) = - x

Expand

2 (3 x^{2} - 1) : 6 x^{2} - 2

6 x^{2} - 2 = - x

Move

x

to the left side

6 x^{2} - 2 + x = 0

Write in the standard form

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

6 x^{2} + x - 2 = 0

Solve with the quadratic formula

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

1^{2} - 4 \cdot 6 (- 2) = 7

x_{1, 2} = \frac{- 1 \pm 7}{2 \cdot 6}

Separate the solutions

x_{1} = \frac{- 1 + 7}{2 \cdot 6}, x_{2} = \frac{- 1 - 7}{2 \cdot 6}

x = \frac{- 1 + 7}{2 \cdot 6} : \frac{1}{2}

x = \frac{- 1 - 7}{2 \cdot 6} : - \frac{2}{3}

The solutions to the quadratic equation are:

x = \frac{1}{2}, x = - \frac{2}{3}

- 2 n^{2} = 14

2 x^{2} - 3 = 19

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

x^{2} + 3 x - 10 = 0

- 4 x^{2} - 12 x + 5 = 0

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