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

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

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

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

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

s = 1 + i, s = 1 - i

Solution steps

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

Solve with the quadratic formula

s_{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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

s = 1 + i, s = 1 - i

5 \leq 3 y + 8 < 17

x^{4} - x^{3} - x^{2} - x - 2

(3 i)^{5}

3 x^{3} - 9 x^{2} - 12 x

(x + 1)^{2} - 81

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