What are the solutions to the equation z^2+16z+89=0 ?

The solutions to the equation z^2+16z+89=0 are z=-8+5i,z=-8-5i

Find the zeros of z^2+16z+89=0

The zeros of z^2+16z+89=0 are z=-8+5i,z=-8-5i

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z^{2} + 16 z + 89 = 0

z = - 8 + 5 i, z = - 8 - 5 i

Solution steps

z^{2} + 16 z + 89 = 0

Solve with the quadratic formula

z_{1, 2} = \frac{- 16 \pm 1 6 ^{2} - 4 \cdot 1 \cdot 89}{2 \cdot 1}

Simplify

1 6^{2} - 4 \cdot 1 \cdot 89 : 10 i

z_{1, 2} = \frac{- 16 \pm 10 i}{2 \cdot 1}

Separate the solutions

z_{1} = \frac{- 16 + 10 i}{2 \cdot 1}, z_{2} = \frac{- 16 - 10 i}{2 \cdot 1}

z = \frac{- 16 + 10 i}{2 \cdot 1} : - 8 + 5 i

z = \frac{- 16 - 10 i}{2 \cdot 1} : - 8 - 5 i

The solutions to the quadratic equation are:

z = - 8 + 5 i, z = - 8 - 5 i

x (x - 9) = 1

(x - 5)^{2} + 7 (x - 5) = - 12

x - 4 + x (x - 3) = 4

4 x^{2} - 1 = - 6 x

3 x^{2} + 14 x + 49 = 2 x^{2} + 100

What are the solutions to the equation z^2+16z+89=0 ?
The solutions to the equation z^2+16z+89=0 are z=-8+5i,z=-8-5i
Find the zeros of z^2+16z+89=0
The zeros of z^2+16z+89=0 are z=-8+5i,z=-8-5i