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

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

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

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

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

x = \frac{1}{2} + i \frac{5}{2}, x = \frac{1}{2} - i \frac{5}{2}

Solution steps

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

Solve with the quadratic formula

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

Simplify

(- 2)^{2} - 4 \cdot 2 \cdot 3 : 25 i

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = \frac{1}{2} + i \frac{5}{2}, x = \frac{1}{2} - i \frac{5}{2}

3 (4 k - \frac{1}{3}) = 12 k - 1

\frac{6}{10} \cdot \frac{4}{9}

\frac{1}{3} + \frac{3}{5}

\frac{9 !}{( 9 - 6 ) !}

x^{2} + 5 x - 90 = 0

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