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

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

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

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

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

x = 1 + 3 i, x = 1 - 3 i

Solution steps

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

Solve with the quadratic formula

x_{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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = 1 + 3 i, x = 1 - 3 i

(x - 1)^{2} = 16

y^{2} + 18 y + 81 = 0

(125 x^{9} y^{6})^{- \frac{5}{3}}

(2 - 5 i) (2 + 5 i)

m > \frac{1}{2} (11 + 18)

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