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

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

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

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

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

z = 1 + 3 i, z = 1 - 3 i

Solution steps

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

Solve with the quadratic formula

z_{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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

z = 1 + 3 i, z = 1 - 3 i

x (x + 8) = 0

4 x^{2} + 5 x - 9 = 0

3 (x - 5)^{2} - 17 = - 5

2 x^{2} + 26 = 20 x

4 x^{2} + 24 x = - 35

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