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

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

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

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

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

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

Solution steps

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

Switch sides

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

Solve with the quadratic formula

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

Simplify

1^{2} - 4 (- 3) (- 4) : 47 i

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

Separate the solutions

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

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

x = \frac{- 1 - 47 i}{2 ( - 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}

4 = 4 - x^{2}

72 + r < - 432

16 + 8

7 - x = - 4

36 y^{4} - 121

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