What are the solutions to the equation 2(3x^2-x)=8-3x ?

The solutions to the equation 2(3x^2-x)=8-3x are x=(-1+sqrt(193))/(12),x=(-1-sqrt(193))/(12)

Find the zeros of 2(3x^2-x)=8-3x

The zeros of 2(3x^2-x)=8-3x are x=(-1+sqrt(193))/(12),x=(-1-sqrt(193))/(12)

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2 (3 x^{2} - x) = 8 - 3 x

x = \frac{- 1 + 193}{12}, x = \frac{- 1 - 193}{12}

Decimal Notation

x = 1.07437 \dots, x = - 1.24103 \dots

Solution steps

2 (3 x^{2} - x) = 8 - 3 x

Expand

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

6 x^{2} - 2 x = 8 - 3 x

Move

3 x

to the left side

6 x^{2} + x = 8

Move

8

to the left side

6 x^{2} + x - 8 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 6 ( - 8 )}{2 \cdot 6}

1^{2} - 4 \cdot 6 (- 8) = 193

x_{1, 2} = \frac{- 1 \pm 193}{2 \cdot 6}

Separate the solutions

x_{1} = \frac{- 1 + 193}{2 \cdot 6}, x_{2} = \frac{- 1 - 193}{2 \cdot 6}

x = \frac{- 1 + 193}{2 \cdot 6} : \frac{- 1 + 193}{12}

x = \frac{- 1 - 193}{2 \cdot 6} : \frac{- 1 - 193}{12}

The solutions to the quadratic equation are:

x = \frac{- 1 + 193}{12}, x = \frac{- 1 - 193}{12}

4 y^{2} = 72

(m + 5) x^{2} + 3 m x + m = 0

3 x^{2} = - 30 x - 27

(h - 5)^{2} = 1

3 x^{2} + 2 x + 160 = 0

What are the solutions to the equation 2(3x^2-x)=8-3x ?
The solutions to the equation 2(3x^2-x)=8-3x are x=(-1+sqrt(193))/(12),x=(-1-sqrt(193))/(12)
Find the zeros of 2(3x^2-x)=8-3x
The zeros of 2(3x^2-x)=8-3x are x=(-1+sqrt(193))/(12),x=(-1-sqrt(193))/(12)