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

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

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

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

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

y = 1 + i \frac{2}{2}, y = 1 - i \frac{2}{2}

Solution steps

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

Solve with the quadratic formula

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

Simplify

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

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

Separate the solutions

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

y = \frac{- ( - 4 ) + 2 2 i}{2 \cdot 2} : 1 + i \frac{2}{2}

y = \frac{- ( - 4 ) - 2 2 i}{2 \cdot 2} : 1 - i \frac{2}{2}

The solutions to the quadratic equation are:

y = 1 + i \frac{2}{2}, y = 1 - i \frac{2}{2}

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

x, f (x + 2) = 2 x^{2} + 3

y (y - 4) = - 8

y, 36 + y^{2} + z^{2} = 4

x, f (x - 2) = x^{2}

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