What are the solutions to the equation z^2+2z+5=0 ?

The solutions to the equation z^2+2z+5=0 are z=-1+2i,z=-1-2i

Find the zeros of z^2+2z+5=0

The zeros of z^2+2z+5=0 are z=-1+2i,z=-1-2i

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

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

Solution steps

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

(7 x^{2}) (x^{3})

5 + 3^{- 2}

\frac{4 ^{5}}{4}

(2 x - 1) (x + 3) < 0

\frac{3}{5} x = - 9

What are the solutions to the equation z^2+2z+5=0 ?
The solutions to the equation z^2+2z+5=0 are z=-1+2i,z=-1-2i
Find the zeros of z^2+2z+5=0
The zeros of z^2+2z+5=0 are z=-1+2i,z=-1-2i