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

The solutions to the equation s^2-2s+4=0 are s=1+sqrt(3)i,s=1-sqrt(3)i

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

The zeros of s^2-2s+4=0 are s=1+sqrt(3)i,s=1-sqrt(3)i

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

s = 1 + 3 i, s = 1 - 3 i

Solution steps

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

Solve with the quadratic formula

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

Simplify

(- 2)^{2} - 4 \cdot 1 \cdot 4 : 23 i

s_{1, 2} = \frac{- ( - 2 ) \pm 2 3 i}{2 \cdot 1}

Separate the solutions

s_{1} = \frac{- ( - 2 ) + 2 3 i}{2 \cdot 1}, s_{2} = \frac{- ( - 2 ) - 2 3 i}{2 \cdot 1}

s = \frac{- ( - 2 ) + 2 3 i}{2 \cdot 1} : 1 + 3 i

s = \frac{- ( - 2 ) - 2 3 i}{2 \cdot 1} : 1 - 3 i

The solutions to the quadratic equation are:

s = 1 + 3 i, s = 1 - 3 i

1 = 3 x^{2}

4 x^{2} - 19 x + 12 = 0

3 x^{2} - 15 x + 186 = 0

2 X^{2} = 0

3 (x^{2} + 1) = 5 (1 - x)

What are the solutions to the equation s^2-2s+4=0 ?
The solutions to the equation s^2-2s+4=0 are s=1+sqrt(3)i,s=1-sqrt(3)i
Find the zeros of s^2-2s+4=0
The zeros of s^2-2s+4=0 are s=1+sqrt(3)i,s=1-sqrt(3)i