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

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

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

The zeros of quadraticformula 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} + 8 x = - 7

(6 - r)^{2} + (3 - r)^{2} = r^{2}

3 t^{2} - 10 t - 6 = 2

9 y^{2} + 1.2 y

5 x^{2} - 32 x - 21 = 0

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