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

The solutions to the equation 3x^2+x=2x-6 are x= 1/6+i(sqrt(71))/6 ,x= 1/6-i(sqrt(71))/6

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

The zeros of 3x^2+x=2x-6 are x= 1/6+i(sqrt(71))/6 ,x= 1/6-i(sqrt(71))/6

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

x = \frac{1}{6} + i \frac{71}{6}, x = \frac{1}{6} - i \frac{71}{6}

Solution steps

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

Move

6

to the left side

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

Move

2 x

to the left side

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

Solve with the quadratic formula

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

Simplify

(- 1)^{2} - 4 \cdot 3 \cdot 6 : 71 i

x_{1, 2} = \frac{- ( - 1 ) \pm 71 i}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- ( - 1 ) + 71 i}{2 \cdot 3}, x_{2} = \frac{- ( - 1 ) - 71 i}{2 \cdot 3}

x = \frac{- ( - 1 ) + 71 i}{2 \cdot 3} : \frac{1}{6} + i \frac{71}{6}

x = \frac{- ( - 1 ) - 71 i}{2 \cdot 3} : \frac{1}{6} - i \frac{71}{6}

The solutions to the quadratic equation are:

x = \frac{1}{6} + i \frac{71}{6}, x = \frac{1}{6} - i \frac{71}{6}

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x + 4 = 2 x

lo g_{4} (48 x + 80 y)

\frac{x - 2}{6} - \frac{x + 1}{3} + \frac{x - 1}{2} = 0

6 x^{2} + 7 x + 2

What are the solutions to the equation 3x^2+x=2x-6 ?
The solutions to the equation 3x^2+x=2x-6 are x= 1/6+i(sqrt(71))/6 ,x= 1/6-i(sqrt(71))/6
Find the zeros of 3x^2+x=2x-6
The zeros of 3x^2+x=2x-6 are x= 1/6+i(sqrt(71))/6 ,x= 1/6-i(sqrt(71))/6