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

The solutions to the equation -3x^2-x-3=0 are x=-1/6-i(sqrt(35))/6 ,x=-1/6+i(sqrt(35))/6

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

The zeros of -3x^2-x-3=0 are x=-1/6-i(sqrt(35))/6 ,x=-1/6+i(sqrt(35))/6

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

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

Solution steps

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

Solve with the quadratic formula

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

Simplify

(- 1)^{2} - 4 (- 3) (- 3) : 35 i

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

Separate the solutions

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

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

x = \frac{- ( - 1 ) - 35 i}{2 ( - 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}

4 (- \frac{1}{2})

\frac{1}{6} - \frac{1}{8}

26 r^{3} s + 52 r^{5} - 39 r^{2} s^{4}

(2 y^{2})^{3}

3 (x + 1) = 2 (x + 1) + 2 x

What are the solutions to the equation -3x^2-x-3=0 ?
The solutions to the equation -3x^2-x-3=0 are x=-1/6-i(sqrt(35))/6 ,x=-1/6+i(sqrt(35))/6
Find the zeros of -3x^2-x-3=0
The zeros of -3x^2-x-3=0 are x=-1/6-i(sqrt(35))/6 ,x=-1/6+i(sqrt(35))/6