What are the solutions to the equation -3x^2+6x-4=0 ?

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

Find the zeros of -3x^2+6x-4=0

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

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- 3 x^{2} + 6 x - 4 = 0

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

Solution steps

- 3 x^{2} + 6 x - 4 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 ( - 3 ) ( - 4 )}{2 ( - 3 )}

Simplify

6^{2} - 4 (- 3) (- 4) : 23 i

x_{1, 2} = \frac{- 6 \pm 2 3 i}{2 ( - 3 )}

Separate the solutions

x_{1} = \frac{- 6 + 2 3 i}{2 ( - 3 )}, x_{2} = \frac{- 6 - 2 3 i}{2 ( - 3 )}

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

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

The solutions to the quadratic equation are:

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

\frac{7}{- i}

2 y^{2} + y - 1 = 0

x^{2} + 10 x + 23 = 0

28 v^{3} + 16 v^{2} - 21 v - 12

4 (5 x + 2) \leq 28

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