What are the solutions to the equation 3x^2-(2x)/3+3=0 ?

The solutions to the equation 3x^2-(2x)/3+3=0 are x= 1/9+i(4sqrt(5))/9 ,x= 1/9-i(4sqrt(5))/9

Find the zeros of 3x^2-(2x)/3+3=0

The zeros of 3x^2-(2x)/3+3=0 are x= 1/9+i(4sqrt(5))/9 ,x= 1/9-i(4sqrt(5))/9

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3 x^{2} - \frac{2 x}{3} + 3 = 0

x = \frac{1}{9} + i \frac{4 5}{9}, x = \frac{1}{9} - i \frac{4 5}{9}

Solution steps

3 x^{2} - \frac{2 x}{3} + 3 = 0

Multiply both sides by

3

9 x^{2} - 2 x + 9 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 9 \cdot 9}{2 \cdot 9}

Simplify

(- 2)^{2} - 4 \cdot 9 \cdot 9 : 85 i

x_{1, 2} = \frac{- ( - 2 ) \pm 8 5 i}{2 \cdot 9}

Separate the solutions

x_{1} = \frac{- ( - 2 ) + 8 5 i}{2 \cdot 9}, x_{2} = \frac{- ( - 2 ) - 8 5 i}{2 \cdot 9}

x = \frac{- ( - 2 ) + 8 5 i}{2 \cdot 9} : \frac{1}{9} + i \frac{4 5}{9}

x = \frac{- ( - 2 ) - 8 5 i}{2 \cdot 9} : \frac{1}{9} - i \frac{4 5}{9}

The solutions to the quadratic equation are:

x = \frac{1}{9} + i \frac{4 5}{9}, x = \frac{1}{9} - i \frac{4 5}{9}

4 (x - 10)^{2} + 20 = 24

16 x^{2} + 48 x + 36 = 0

- x^{2} = 64

x^{2} - 2 x - 26 = 0

0 = 3 x^{2} - 6 x - 9

What are the solutions to the equation 3x^2-(2x)/3+3=0 ?
The solutions to the equation 3x^2-(2x)/3+3=0 are x= 1/9+i(4sqrt(5))/9 ,x= 1/9-i(4sqrt(5))/9
Find the zeros of 3x^2-(2x)/3+3=0
The zeros of 3x^2-(2x)/3+3=0 are x= 1/9+i(4sqrt(5))/9 ,x= 1/9-i(4sqrt(5))/9