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

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

Find the zeros of roots s^2-2s+2

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

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roots

s^{2} - 2 s + 2

s = 1 + i, s = 1 - i

Solution steps

s^{2} - 2 s + 2

The roots are the intercepts with the x-axis

(y = 0)

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

2 x^{2} + 26 = 20 x

y, 2 x^{2} + 3 x y + y^{2} = 5

(6 b - 8)^{2} - 19 = 0

8 x^{2} - 2 x = 15

3 x^{2} - 15 = 33

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