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

x^{2} + (\frac{5}{6})^{2} = 1

s^{2} - 49 = 0

x, 4 x^{2} + 9 (m x - 1)^{2} = 5

12 x^{2} - 2028 x + 62370 = 0

2 5^{2} = 7^{2} + b^{2}

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