What are the solutions to the equation z^2+2z+17=0 ?

The solutions to the equation z^2+2z+17=0 are z=-1+4i,z=-1-4i

Find the zeros of z^2+2z+17=0

The zeros of z^2+2z+17=0 are z=-1+4i,z=-1-4i

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z^{2} + 2 z + 17 = 0

z = - 1 + 4 i, z = - 1 - 4 i

Solution steps

z^{2} + 2 z + 17 = 0

Solve with the quadratic formula

z_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 1 \cdot 17}{2 \cdot 1}

Simplify

2^{2} - 4 \cdot 1 \cdot 17 : 8 i

z_{1, 2} = \frac{- 2 \pm 8 i}{2 \cdot 1}

Separate the solutions

z_{1} = \frac{- 2 + 8 i}{2 \cdot 1}, z_{2} = \frac{- 2 - 8 i}{2 \cdot 1}

z = \frac{- 2 + 8 i}{2 \cdot 1} : - 1 + 4 i

z = \frac{- 2 - 8 i}{2 \cdot 1} : - 1 - 4 i

The solutions to the quadratic equation are:

z = - 1 + 4 i, z = - 1 - 4 i

x, x^{2} + y^{2} + z^{2} = 4 x

x, x^{2} + y^{2} - z = 0

5 y^{2} - 12 y + 2 = - 3 y

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

7 x^{2} = - 14 x

What are the solutions to the equation z^2+2z+17=0 ?
The solutions to the equation z^2+2z+17=0 are z=-1+4i,z=-1-4i
Find the zeros of z^2+2z+17=0
The zeros of z^2+2z+17=0 are z=-1+4i,z=-1-4i