What are the solutions to the equation 2x^2-4x+16=0 ?

The solutions to the equation 2x^2-4x+16=0 are x=1+sqrt(7)i,x=1-sqrt(7)i

Find the zeros of 2x^2-4x+16=0

The zeros of 2x^2-4x+16=0 are x=1+sqrt(7)i,x=1-sqrt(7)i

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2 x^{2} - 4 x + 16 = 0

x = 1 + 7 i, x = 1 - 7 i

Solution steps

2 x^{2} - 4 x + 16 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 2 \cdot 16}{2 \cdot 2}

Simplify

(- 4)^{2} - 4 \cdot 2 \cdot 16 : 47 i

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

Separate the solutions

x_{1} = \frac{- ( - 4 ) + 4 7 i}{2 \cdot 2}, x_{2} = \frac{- ( - 4 ) - 4 7 i}{2 \cdot 2}

x = \frac{- ( - 4 ) + 4 7 i}{2 \cdot 2} : 1 + 7 i

x = \frac{- ( - 4 ) - 4 7 i}{2 \cdot 2} : 1 - 7 i

The solutions to the quadratic equation are:

x = 1 + 7 i, x = 1 - 7 i

x^{2} - 2 x - 15 = - x + 5

x^{2} + 17 x + 72 = 12

x^{2} = 1 2^{2} + 1 6^{2}

4 z^{2} + 16 z + 16 = 0

17 - 4 x = 16 x^{2} + 24 x - 8

What are the solutions to the equation 2x^2-4x+16=0 ?
The solutions to the equation 2x^2-4x+16=0 are x=1+sqrt(7)i,x=1-sqrt(7)i
Find the zeros of 2x^2-4x+16=0
The zeros of 2x^2-4x+16=0 are x=1+sqrt(7)i,x=1-sqrt(7)i