What are the solutions to the equation 2x^2-4x+8=0 ?

The solutions to the equation 2x^2-4x+8=0 are x=1+sqrt(3)i,x=1-sqrt(3)i

Find the zeros of 2x^2-4x+8=0

The zeros of 2x^2-4x+8=0 are x=1+sqrt(3)i,x=1-sqrt(3)i

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2 x^{2} - 4 x + 8 = 0

x = 1 + 3 i, x = 1 - 3 i

Solution steps

2 x^{2} - 4 x + 8 = 0

Solve with the quadratic formula

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

Simplify

(- 4)^{2} - 4 \cdot 2 \cdot 8 : 43 i

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

Separate the solutions

x_{1} = \frac{- ( - 4 ) + 4 3 i}{2 \cdot 2}, x_{2} = \frac{- ( - 4 ) - 4 3 i}{2 \cdot 2}

x = \frac{- ( - 4 ) + 4 3 i}{2 \cdot 2} : 1 + 3 i

x = \frac{- ( - 4 ) - 4 3 i}{2 \cdot 2} : 1 - 3 i

The solutions to the quadratic equation are:

x = 1 + 3 i, x = 1 - 3 i

x^{2} + \frac{1}{3} x + 5

- 2 x^{2} + 6 = 0

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

p^{2} - 16 p + 13 = - 3 p^{2}

(3 x - 4) (x + 5) = 0

What are the solutions to the equation 2x^2-4x+8=0 ?
The solutions to the equation 2x^2-4x+8=0 are x=1+sqrt(3)i,x=1-sqrt(3)i
Find the zeros of 2x^2-4x+8=0
The zeros of 2x^2-4x+8=0 are x=1+sqrt(3)i,x=1-sqrt(3)i