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

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

Find the zeros of k^2-2k+2=0

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

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k^{2} - 2 k + 2 = 0

k = 1 + i, k = 1 - i

Solution steps

k^{2} - 2 k + 2 = 0

Solve with the quadratic formula

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

Simplify

(- 2)^{2} - 4 \cdot 1 \cdot 2 : 2 i

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

k = 1 + i, k = 1 - i

(12 x^{2} + 2 x - 2) = 0

6 x^{2} = 4 x - 1

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

- 6 x^{2} + 5 = - 145

9 x^{2} - 54 x + 85 = 0

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