What are the solutions to the equation 3+x-3x^2=9 ?

The solutions to the equation 3+x-3x^2=9 are x= 1/6-i(sqrt(71))/6 ,x= 1/6+i(sqrt(71))/6

Find the zeros of 3+x-3x^2=9

The zeros of 3+x-3x^2=9 are x= 1/6-i(sqrt(71))/6 ,x= 1/6+i(sqrt(71))/6

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3 + x - 3 x^{2} = 9

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

Solution steps

3 + x - 3 x^{2} = 9

Move

9

to the left side

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

Solve with the quadratic formula

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

Simplify

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

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

Separate the solutions

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

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

x = \frac{- 1 - 71 i}{2 ( - 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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\frac{1}{5} \cdot \frac{1}{5} \cdot \frac{1}{5} \cdot \frac{1}{5}

2 x^{2} + 16 x + 30 = 0

12 y + 24

2 x^{2} + 2 = 5 x

What are the solutions to the equation 3+x-3x^2=9 ?
The solutions to the equation 3+x-3x^2=9 are x= 1/6-i(sqrt(71))/6 ,x= 1/6+i(sqrt(71))/6
Find the zeros of 3+x-3x^2=9
The zeros of 3+x-3x^2=9 are x= 1/6-i(sqrt(71))/6 ,x= 1/6+i(sqrt(71))/6