What are the solutions to the equation z(z-2)=-5 ?

The solutions to the equation z(z-2)=-5 are z=1+2i,z=1-2i

Find the zeros of z(z-2)=-5

The zeros of z(z-2)=-5 are z=1+2i,z=1-2i

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z (z - 2) = - 5

z = 1 + 2 i, z = 1 - 2 i

Solution steps

z (z - 2) = - 5

Expand

z (z - 2) : z^{2} - 2 z

z^{2} - 2 z = - 5

Move

5

to the left side

z^{2} - 2 z + 5 = 0

Solve with the quadratic formula

z_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 1 \cdot 5}{2 \cdot 1}

Simplify

(- 2)^{2} - 4 \cdot 1 \cdot 5 : 4 i

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

z = 1 + 2 i, z = 1 - 2 i

(x + 3) (x + 3) = 1

y^{2} + 12 y + 36 = 0

x^{2} + 5 x + 1.25 = 0

(x + 2)^{2} = 2 x + 2

3 a^{2} - 5 a - 2 = 0

What are the solutions to the equation z(z-2)=-5 ?
The solutions to the equation z(z-2)=-5 are z=1+2i,z=1-2i
Find the zeros of z(z-2)=-5
The zeros of z(z-2)=-5 are z=1+2i,z=1-2i