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

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

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

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

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

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

Solution steps

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

Solve with the quadratic formula

x_{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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

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

x^{2} - 2.456 x + 1.25 = 0

2 x^{2} - 5 x + 5 = 0

ln (e^{\frac{1}{4}})

12 (x + 7) = - 3 (x + 2)

- 1 (b - 9) < 6

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