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

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

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

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

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

y = \frac{2}{3} + i \frac{2}{3}, y = \frac{2}{3} - i \frac{2}{3}

Solution steps

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

Solve with the quadratic formula

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

Simplify

(- 4)^{2} - 4 \cdot 3 \cdot 2 : 22 i

y_{1, 2} = \frac{- ( - 4 ) \pm 2 2 i}{2 \cdot 3}

Separate the solutions

y_{1} = \frac{- ( - 4 ) + 2 2 i}{2 \cdot 3}, y_{2} = \frac{- ( - 4 ) - 2 2 i}{2 \cdot 3}

y = \frac{- ( - 4 ) + 2 2 i}{2 \cdot 3} : \frac{2}{3} + i \frac{2}{3}

y = \frac{- ( - 4 ) - 2 2 i}{2 \cdot 3} : \frac{2}{3} - i \frac{2}{3}

The solutions to the quadratic equation are:

y = \frac{2}{3} + i \frac{2}{3}, y = \frac{2}{3} - i \frac{2}{3}

x^{2} - 10 x - 7 = 3

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

x^{2} + 9 = 2 x (3 - x) + 9

x, f (x + 1) = 2 x^{2}

r^{2} - 10 r = - 17

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