What are the solutions to the equation -3x^2+x-5=0 ?

The solutions to the equation -3x^2+x-5=0 are x= 1/6-i(sqrt(59))/6 ,x= 1/6+i(sqrt(59))/6

Find the zeros of -3x^2+x-5=0

The zeros of -3x^2+x-5=0 are x= 1/6-i(sqrt(59))/6 ,x= 1/6+i(sqrt(59))/6

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- 3 x^{2} + x - 5 = 0

x = \frac{1}{6} - i \frac{59}{6}, x = \frac{1}{6} + i \frac{59}{6}

Solution steps

- 3 x^{2} + x - 5 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 ( - 3 ) ( - 5 )}{2 ( - 3 )}

Simplify

1^{2} - 4 (- 3) (- 5) : 59 i

x_{1, 2} = \frac{- 1 \pm 59 i}{2 ( - 3 )}

Separate the solutions

x_{1} = \frac{- 1 + 59 i}{2 ( - 3 )}, x_{2} = \frac{- 1 - 59 i}{2 ( - 3 )}

x = \frac{- 1 + 59 i}{2 ( - 3 )} : \frac{1}{6} - i \frac{59}{6}

x = \frac{- 1 - 59 i}{2 ( - 3 )} : \frac{1}{6} + i \frac{59}{6}

The solutions to the quadratic equation are:

x = \frac{1}{6} - i \frac{59}{6}, x = \frac{1}{6} + i \frac{59}{6}

\frac{10 x + 4}{7} = 8

3 x^{2} < - 4 x + 4

x^{3} - 10 x^{2} + 41 x

(\frac{5 x ^{3} y}{20 x y ^{5}})^{4}

64 + x^{3}

What are the solutions to the equation -3x^2+x-5=0 ?
The solutions to the equation -3x^2+x-5=0 are x= 1/6-i(sqrt(59))/6 ,x= 1/6+i(sqrt(59))/6
Find the zeros of -3x^2+x-5=0
The zeros of -3x^2+x-5=0 are x= 1/6-i(sqrt(59))/6 ,x= 1/6+i(sqrt(59))/6