What are the solutions to the equation z^2-4z+8=0 ?

The solutions to the equation z^2-4z+8=0 are z=2+2i,z=2-2i

Find the zeros of z^2-4z+8=0

The zeros of z^2-4z+8=0 are z=2+2i,z=2-2i

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z^{2} - 4 z + 8 = 0

z = 2 + 2 i, z = 2 - 2 i

Solution steps

z^{2} - 4 z + 8 = 0

Solve with the quadratic formula

z_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 1 \cdot 8}{2 \cdot 1}

Simplify

(- 4)^{2} - 4 \cdot 1 \cdot 8 : 4 i

z_{1, 2} = \frac{- ( - 4 ) \pm 4 i}{2 \cdot 1}

Separate the solutions

z_{1} = \frac{- ( - 4 ) + 4 i}{2 \cdot 1}, z_{2} = \frac{- ( - 4 ) - 4 i}{2 \cdot 1}

z = \frac{- ( - 4 ) + 4 i}{2 \cdot 1} : 2 + 2 i

z = \frac{- ( - 4 ) - 4 i}{2 \cdot 1} : 2 - 2 i

The solutions to the quadratic equation are:

z = 2 + 2 i, z = 2 - 2 i

(n + 2) x^{2} - 6 n x + 9 = 0

x^{2} + 14 x + 38 = 0

x^{2} - (m + 7) x + 25 = 0

15 x^{2} - 31 x + 10 = 0

2 w^{2} + 29 = - 21

What are the solutions to the equation z^2-4z+8=0 ?
The solutions to the equation z^2-4z+8=0 are z=2+2i,z=2-2i
Find the zeros of z^2-4z+8=0
The zeros of z^2-4z+8=0 are z=2+2i,z=2-2i