What are the solutions to the equation 2x^2-6x+7=0 ?

The solutions to the equation 2x^2-6x+7=0 are x= 3/2+i(sqrt(5))/2 ,x= 3/2-i(sqrt(5))/2

Find the zeros of 2x^2-6x+7=0

The zeros of 2x^2-6x+7=0 are x= 3/2+i(sqrt(5))/2 ,x= 3/2-i(sqrt(5))/2

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2 x^{2} - 6 x + 7 = 0

x = \frac{3}{2} + i \frac{5}{2}, x = \frac{3}{2} - i \frac{5}{2}

Solution steps

2 x^{2} - 6 x + 7 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 6 ) \pm ( - 6 ) ^{2} - 4 \cdot 2 \cdot 7}{2 \cdot 2}

Simplify

(- 6)^{2} - 4 \cdot 2 \cdot 7 : 25 i

x_{1, 2} = \frac{- ( - 6 ) \pm 2 5 i}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 6 ) + 2 5 i}{2 \cdot 2}, x_{2} = \frac{- ( - 6 ) - 2 5 i}{2 \cdot 2}

x = \frac{- ( - 6 ) + 2 5 i}{2 \cdot 2} : \frac{3}{2} + i \frac{5}{2}

x = \frac{- ( - 6 ) - 2 5 i}{2 \cdot 2} : \frac{3}{2} - i \frac{5}{2}

The solutions to the quadratic equation are:

x = \frac{3}{2} + i \frac{5}{2}, x = \frac{3}{2} - i \frac{5}{2}

k^{4} - 4 k^{2} + 8 k^{3} - 32 k + 12 k^{2} - 48

28 (10 - \frac{28}{3.5}) = x

1027

3 x + 7 = - 8

(z y x)^{8}

What are the solutions to the equation 2x^2-6x+7=0 ?
The solutions to the equation 2x^2-6x+7=0 are x= 3/2+i(sqrt(5))/2 ,x= 3/2-i(sqrt(5))/2
Find the zeros of 2x^2-6x+7=0
The zeros of 2x^2-6x+7=0 are x= 3/2+i(sqrt(5))/2 ,x= 3/2-i(sqrt(5))/2