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

(\frac{3}{4})^{- 7}

∣ 2 x - 5 ∣ > 25

x^{2} \leq 6 x - 7

\frac{1}{3} (3)

(\frac{8 x ^{4}}{2 x ^{6}})^{5}

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