What are the solutions to the equation 3x^2+x+12=9 ?

The solutions to the equation 3x^2+x+12=9 are x=-1/6+i(sqrt(35))/6 ,x=-1/6-i(sqrt(35))/6

Find the zeros of 3x^2+x+12=9

The zeros of 3x^2+x+12=9 are x=-1/6+i(sqrt(35))/6 ,x=-1/6-i(sqrt(35))/6

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3 x^{2} + x + 12 = 9

x = - \frac{1}{6} + i \frac{35}{6}, x = - \frac{1}{6} - i \frac{35}{6}

Solution steps

3 x^{2} + x + 12 = 9

Move

9

to the left side

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

Solve with the quadratic formula

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

Simplify

1^{2} - 4 \cdot 3 \cdot 3 : 35 i

x_{1, 2} = \frac{- 1 \pm 35 i}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- 1 + 35 i}{2 \cdot 3}, x_{2} = \frac{- 1 - 35 i}{2 \cdot 3}

x = \frac{- 1 + 35 i}{2 \cdot 3} : - \frac{1}{6} + i \frac{35}{6}

x = \frac{- 1 - 35 i}{2 \cdot 3} : - \frac{1}{6} - i \frac{35}{6}

The solutions to the quadratic equation are:

x = - \frac{1}{6} + i \frac{35}{6}, x = - \frac{1}{6} - i \frac{35}{6}

0 - 1

(a - 2)^{2} = - 5

3 x^{2} + 16 x + 5

x^{2} - 225 = 0

- 6 x^{2} + 3 x + 13 = - 2 x^{2}

What are the solutions to the equation 3x^2+x+12=9 ?
The solutions to the equation 3x^2+x+12=9 are x=-1/6+i(sqrt(35))/6 ,x=-1/6-i(sqrt(35))/6
Find the zeros of 3x^2+x+12=9
The zeros of 3x^2+x+12=9 are x=-1/6+i(sqrt(35))/6 ,x=-1/6-i(sqrt(35))/6