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

The solutions to the equation 2x^2-4x+4=0 are x=1+i,x=1-i

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

The zeros of 2x^2-4x+4=0 are x=1+i,x=1-i

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

x = 1 + i, x = 1 - i

Solution steps

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

Solve with the quadratic formula

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

Simplify

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = 1 + i, x = 1 - i

(5)^{\frac{3}{2}}

x^{2} - 9 \leq 0

3 x^{2} - 6 = 7 x

15 - \frac{1}{2} 2

3^{4 x - 7} < 4^{2 x + 3}

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