What are the solutions to the equation 3x^2+x+4=0 ?

The solutions to the equation 3x^2+x+4=0 are x=-1/6+i(sqrt(47))/6 ,x=-1/6-i(sqrt(47))/6

Find the zeros of 3x^2+x+4=0

The zeros of 3x^2+x+4=0 are x=-1/6+i(sqrt(47))/6 ,x=-1/6-i(sqrt(47))/6

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3 x^{2} + x + 4 = 0

x = - \frac{1}{6} + i \frac{47}{6}, x = - \frac{1}{6} - i \frac{47}{6}

Solution steps

3 x^{2} + x + 4 = 0

Solve with the quadratic formula

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

Simplify

1^{2} - 4 \cdot 3 \cdot 4 : 47 i

x_{1, 2} = \frac{- 1 \pm 47 i}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- 1 + 47 i}{2 \cdot 3}, x_{2} = \frac{- 1 - 47 i}{2 \cdot 3}

x = \frac{- 1 + 47 i}{2 \cdot 3} : - \frac{1}{6} + i \frac{47}{6}

x = \frac{- 1 - 47 i}{2 \cdot 3} : - \frac{1}{6} - i \frac{47}{6}

The solutions to the quadratic equation are:

x = - \frac{1}{6} + i \frac{47}{6}, x = - \frac{1}{6} - i \frac{47}{6}

\frac{7}{4 3}

3 x^{2} - 1 = 7 x

8 \cdot 14

6 u^{3} v^{3} + 18 u^{2} v w^{4} - 12 u^{2} v^{3}

\frac{( 18 + 2 ) ^{2}}{8 - 2}

What are the solutions to the equation 3x^2+x+4=0 ?
The solutions to the equation 3x^2+x+4=0 are x=-1/6+i(sqrt(47))/6 ,x=-1/6-i(sqrt(47))/6
Find the zeros of 3x^2+x+4=0
The zeros of 3x^2+x+4=0 are x=-1/6+i(sqrt(47))/6 ,x=-1/6-i(sqrt(47))/6