What are the solutions to the equation 2x^2-2x+8=0 ?

The solutions to the equation 2x^2-2x+8=0 are x= 1/2+i(sqrt(15))/2 ,x= 1/2-i(sqrt(15))/2

Find the zeros of 2x^2-2x+8=0

The zeros of 2x^2-2x+8=0 are x= 1/2+i(sqrt(15))/2 ,x= 1/2-i(sqrt(15))/2

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2 x^{2} - 2 x + 8 = 0

x = \frac{1}{2} + i \frac{15}{2}, x = \frac{1}{2} - i \frac{15}{2}

Solution steps

2 x^{2} - 2 x + 8 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 2 \cdot 8}{2 \cdot 2}

Simplify

(- 2)^{2} - 4 \cdot 2 \cdot 8 : 215 i

x_{1, 2} = \frac{- ( - 2 ) \pm 2 15 i}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 2 ) + 2 15 i}{2 \cdot 2}, x_{2} = \frac{- ( - 2 ) - 2 15 i}{2 \cdot 2}

x = \frac{- ( - 2 ) + 2 15 i}{2 \cdot 2} : \frac{1}{2} + i \frac{15}{2}

x = \frac{- ( - 2 ) - 2 15 i}{2 \cdot 2} : \frac{1}{2} - i \frac{15}{2}

The solutions to the quadratic equation are:

x = \frac{1}{2} + i \frac{15}{2}, x = \frac{1}{2} - i \frac{15}{2}

x, y^{2} (x^{2} - 4) = x + 2

x^{2} + 4 = 20

2 x^{2} - 3 x - 5 = - 10 x

x^{2} - 2 x + 17

e^{2} = 0

What are the solutions to the equation 2x^2-2x+8=0 ?
The solutions to the equation 2x^2-2x+8=0 are x= 1/2+i(sqrt(15))/2 ,x= 1/2-i(sqrt(15))/2
Find the zeros of 2x^2-2x+8=0
The zeros of 2x^2-2x+8=0 are x= 1/2+i(sqrt(15))/2 ,x= 1/2-i(sqrt(15))/2