What are the solutions to the equation 3x^2+6x+7=0 ?

The solutions to the equation 3x^2+6x+7=0 are x=-1+i(2sqrt(3))/3 ,x=-1-i(2sqrt(3))/3

Find the zeros of 3x^2+6x+7=0

The zeros of 3x^2+6x+7=0 are x=-1+i(2sqrt(3))/3 ,x=-1-i(2sqrt(3))/3

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3 x^{2} + 6 x + 7 = 0

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

Solution steps

3 x^{2} + 6 x + 7 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 \cdot 3 \cdot 7}{2 \cdot 3}

Simplify

6^{2} - 4 \cdot 3 \cdot 7 : 43 i

x_{1, 2} = \frac{- 6 \pm 4 3 i}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- 6 + 4 3 i}{2 \cdot 3}, x_{2} = \frac{- 6 - 4 3 i}{2 \cdot 3}

x = \frac{- 6 + 4 3 i}{2 \cdot 3} : - 1 + i \frac{2 3}{3}

x = \frac{- 6 - 4 3 i}{2 \cdot 3} : - 1 - i \frac{2 3}{3}

The solutions to the quadratic equation are:

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

4 x^{4} - 25 x^{2} + 36

\frac{x}{8} = \frac{3}{5}

3 x^{2} + 15 x

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

x^{- \frac{1}{2}} \cdot x^{- 1}

What are the solutions to the equation 3x^2+6x+7=0 ?
The solutions to the equation 3x^2+6x+7=0 are x=-1+i(2sqrt(3))/3 ,x=-1-i(2sqrt(3))/3
Find the zeros of 3x^2+6x+7=0
The zeros of 3x^2+6x+7=0 are x=-1+i(2sqrt(3))/3 ,x=-1-i(2sqrt(3))/3